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ELEMENTARY    TBEATISE 


STATICS, 


BY    GASPARD    MONGE. 


ft  !J3ingnt{itjinil  Sntia  nf  tljB  Slutljnt 


TRANSLATED  FROM  TUB  FRENCH, 

BY  WOODS    BAKER,    A.M. 

OF  THE  UNITED  STATES  COAST   SURVEY. 


PHILADELPHIA: 

E.  C.  &  J.  BIDDLE,  No.  6  S.  FIFTH  STREET. 
1851. 


ENTERED  according  to  Act  of  Congress,  in  the  year  1851,  by 

E.  C.  &  J.  BIDDLE, 

in  the  Clerk's  Office  of  the  District  Court  of  the  United  States,  in  and  for  the 
Eastern  District  of  Pennsylvania. 


STEREOTYPED   BY   L.   JOHNSON  AND   CO. 

PHILADELPHIA. 
PRINTED  BY   T.    K.    AND   P.   G.    COLLINS. 


PREFACE 


AMERICAN  EDITION. 


A  WANT  has  long  been  felt  in  this  country  of  a  good 
elementary  treatise  on  Theoretical  Mechanics.  The 
books  on  that  subject,  in  the  English  language,  are 
mostly  voluminous,  and  either  adapted  to  the  compre 
hension  of  those  only  who  have  mastered  the  various 
branches  of  analytical  mathematics,  or  composed  chiefly 
of  practical  and  descriptive  details.  Hence,  an  accurate 
knowledge  of  the  general  laws,  or  grand  fundamental 
truths  of  mechanics,  so  important  to  all  men,  in  this 
eminently  practical  age  and  country,  and  especially  to 
those  who  have  some  one  of  the  useful  arts  as  their 
daily  occupation,  has  hitherto  been  attainable  only  by 
highly  educated  men. 

One  of  the  consequences  of  the  want  of  a  familiar 
acquaintance  with  the  mechanical  laws,  upon  which  all 
machines,  of  whatever  possible  kind,  must  depend,  is 
the  large  number  of  failures  of  inventions  occurring 
every  year.  The  authors  of  such  attempts  generally 

3 


4  PREFACE. 

have  mental  ingenuity  enough,  but,  unfortunately,  they 
have  not  the  knowledge  necessary  to  render  their  con 
trivances  consistent  with  the  laws  of  Nature,  or  adapted 
to  attain  the  proposed  ends  by  the  best  possible  means. 
Machines,  deficient  in  either  of  these  essential  matters, 
must  sooner  or  later  be  discarded;  and  their  disap 
pointed  inventors  have  then  to  regret  the  loss  of  their 
money  and  time,  which  proper  information  would  have 
prevented. 

To  supply  this  urgent  need,  in  part,  the  following 
little  book  upon  Statics,  or  the  science  which  treats  of 
the  equilibrium  of  forces  applied  to  solid  bodies,  has 
been  translated.  It  has  long  been  known  and  highly 
admired  by  those  who  are  familiar  with  the  scientific 
literature  of  France ;  but  to  persons  who  have  little  or 
no  acquaintance  with  French  authors  on  such  subjects, 
it  may  be  well  to  mention,  that  correct  information,  so 
well  digested,  precise  and  clear,  can  be  obtained  from 
the  literature  of  no  other  nation  or  language. 

From  the  advertisement  prefixed  to  the  seventh 
French  edition  of  this  treatise,  which  has  gone  through 
eight  editions  in  France  aloner  besides  several  that  have 
been  published  at  Brussels,  the  following  extracts  are 
made : 

"  The  first  edition  of  this  work,  which  appeared  in 
1786,  was  specially  intended  for  young  candidates 


PREFACE.  5 

for  the  Navy;  now,  it  is  one  of  the  standard  books 
most  generally  followed.  A  correct  and  clear  style, 
rigorous  demonstrations,  and  well  connected  proposi 
tions,  have  long  caused  it  to  be  preferred  for  instruction 
in  Statics.  It  is  the  first  book  in  which  there  has  been 
collected  all  that  can  be  demonstrated  in  Statics  syn 
thetically.  After  having  studied  Euclid's  Geometry, 
this  work  will  be  read  without  difficulty.  Being  mo 
delled  upon  the  method  of  the  ancient  geometers,  it 
presents  very  clear  ideas  upon  an  abstract  science,  of 
which  a  great  number  of  useful  applications  are  made." 

"  A  profound  knowledge  of  Statics  requires  the  aid 
of  mathematical  analysis,  that  is  to  say,  of  Algebra,  and 
the  Differential  Calculus ;  but  it  is  as  important  for  be 
ginners  to  study  synthetical  statics  before  analytical 
statics,  as  it  is  fit  to  precede  the  study  of  analytical 
geometry  by  that  of  elementary  geometry." 

"  The  discussion  of  the  elementary  machines  offers  to 
M.  Monge  an  occasion  for  showing  the  truth  of  a  prin 
ciple,  which  the  author  of  the  Mecanique  Analytique 
(Lagrange)  has  rendered  so  productive,  and  which  is 
known  as  the  principle  of  virtual  velocities." 

The  treatise  on  Statics,  of  Monge,  is  a  necessary 
introduction  to  the  work  of  Poisson,  which  is  a  large 
and  thorough  analytical  treatise,  composed  for  the 
pupils  of  the  Polytechnic  School  of  Paris. 


D  PREFACE. 

The  translator  would  here  acknowledge  his  obligation 
to  his  friends,  R.  S.  McCulloh,  Esq.,  Professor  of  Na 
tural  Philosophy  in  the  College  of  New  Jersey,  and 
J.  B.  Reynolds,  Esq.,  Engineer  and  Lecturer  on  Me 
chanical  Philosophy  before  the  Franklin  Institute,  Phila 
delphia,  for  their  kindness  in  giving  him  valuable  advice 
and  assistance  in  preparing  this  work  for  publication. 

Washington,  Nov.  1850. 


BIOGRAPHICAL    NOTICE 


OP 


G  A  SPAED  MONGE. 


IT  may  be  neither  uninteresting  nor  out  of  place  here,  to 
give  a  slight  sketch  of  the  life  of  the  author  of  this  work, 
who  was  one  of  the  most  distinguished  of  the  eminent  men  of 
science  in  France  at  the  close  of  the  last  century.  It  is 
selected  from  two  papers,*  written  by  his  pupils,  MM.  Brisson 
and  Dupin,  and  presented  by  M.  Delambrc  in  his  Analysis  of 
the  Labors  of  the  Academy  of  Sciences  during  the  year  1818. 

Graspard  Monge  was  born  in  1746 His  progress 

was  so  meritorious  as  to  procure  for  him  the  Professorship  of 
Physics  in  the  College  of  Lyons,  the  year  after  he  had  com 
pleted  his  studies  in  that  institution Having 

gone  to  Beaune  during  vacation,  he  undertook  to  draw  the 
plan  of  that  city ;  for  which  purpose  he  was  obliged  to  make 
the  necessary  instruments.  He  presented  his  labor  to  the 
rulers  of  his  native  city,  who  compensated  the  young  author 
as  generously  as  the  limited  means  of  the  public  purse  would 


*  Notice  Historique  sur  Gaspard  Monge,  by  M.  Brisson ;  and  Essai  Historique  sur  Us 
Set-vices  et  les  Travaux  Scientifiques  de  G.  Mange,  by  M.  Dupin.  Sec  the  advertisement 
to  the  7th  edition  of  G&ometrie  Descriptive,  par  G.  Mongc,  Paris,  1847. 

7 


8  BIOGRAPHICAL    NOTICE. 

permit.  A  lieutenant-colonel  of  military  engineers,  who  was 
then  at  Beaune,  obtained  for  Monge  the  position  of  draughts 
man  and  pupil  in  the  school  for  engineers,  and  conductors  of 

works  of  fortification As  he  designed  with  rare 

ability,  his  manual  talent  only  was  regarded ,  but  he  already 
felt  his  strength,  and  could  not  think,  without  indignation,  of 
the  exclusive  esteem  accorded  to  his  mechanical  skill.  He 
said  long  afterwards :  "  I  was  tempted  a  thousand  times  to 
tear  my  drawings  to  pieces  through  spite  at  the  ado  which 
was  made  about  them,  as  though  I  had  no  ability  for  produc 
ing  any  thing  else."  The  director  of  the  school  caused  him 
to  make  some  practical  calculations  of  a  particular  case  of  de 
filement,  an  operation  serving  to  combine  the  relief  and  trace 
of  a  fortification,  so  that  the  defender  may  be  sheltered  from 
the  fire  of  the  assailant.  Monge  abandoned  the  mode  pre 
viously  followed,  and  discovered  the  first  geometric  and  general 
method  that  has  been  given  for  this  important  operation.  .  . 
....  By  applying  successively  his  mathematical  talent  to 
different  questions  of  an  analogous  kind,  and  generalizing  his 
means  of  conceiving  and  operating,  he  at  length  succeeded  in 
forming  a  body  of  doctrine,  which  was  his  Descriptive  Ge 
ometry For  more  than  twenty  years  he  found  it 

impossible  to  get  the  application  of  his  geometry  to  draughts 
of  carpentry  taught  to  the  corps  at  Mezieres.  He  was  more 
fortunate,  however,  in  the  application  to  stone  cutting;  he 
carefully  followed  the  methods  employed  in  this  art,  and  im 
proved  them  by  rendering  them  more  simple  by  means  of  his 
geometry. 

His  scientific  labors  secured  him  the  nomination  of  Lecturer 
on  Mathematics  and  Physics,  to  succeed  Nollet  and  Bossut. 
He  was  afterwards  appointed  the  regular  professor,  when, 
turning  his  attention  to  a  series  of  natural  phenomena,  he 
made  numerous  experiments  upon  electricity,  explained  the 
phenomena  relating  to  capillarity,  and  was  the  founder  of  an 


BIOGRAPHICAL   NOTICE.  9 

ingenious  system  of  meteorology.  He  effected  the  decomposi 
tion  of  water,  and  arrived  at  this  great  discovery  without 
having  had  any  knowledge  of  the  researches  made  a  short 
time  before  by  Lavoisier,  Laplace,  and  Cavendish.  He  was 
not  satisfied  with  explaining  the  theories  of  science  and  their 
application  to  the  students  in  the  lecture  rooms ;  but  liked  to 
conduct  his  pupils  wherever  the  phenomena  of  nature  and  the 
works  of  art  could  render  these  applications  interesting  and 
apparent  to  the  senses.  He  infused  into  his  pupils  his  own 
ardor  and  enthusiasm,  and  rendered  delightful,  observations 
and  investigations,  which,  had  they  been  viewed  only  abstractly 
in  the  narrow  precincts  of  a  lecture-room,  would  have  appeared 
a  wearisome  study. 

In  1780,  to  attract  Monge  to  Paris,  he  was  associated  with 
Bossut  as  Professor  of  the  course  of  Hydrodynamics,  estab 
lished  by  Turgot.  In  order  to  reconcile  the  duties  of  the 
two  places  he  had  to  fulfil,  he  passed  six  months  of  the  year 
at  Mezieres,  and  six  months  at  Paris.  The  same  year  he  was 
elected  a  member  of  the  Academy  of  Sciences;  and,  at  the 
death  of  Bezout,  in  1783,  was  chosen  to  replace  that  cele 
brated  examiner  for  the  Navy.  The  Marquis  de  Castries  fre 
quently  requested  Monge  to  re-write  the  Elementary  Course 
of  Mathematics  for  the  pupils  of  the  Navy ;  but  he  always 
declined.  "Bezout  has  left,"  said  Monge,  "a  widow  who  has 
no  other  fortune  than  the  writings  of  her  husband,  and  I  can 
not  think  of  snatching  away  the  bread  from  the  wife  of  a 
man  who  has  rendered  such  important  service  to  science  and 
the  country."  The  only  elementary  work  which  Monge  pub 
lished,  was  his  Treatise  on  Statics ;  and,  with  the  exception 
of  a  few  passages,  where  the  evidence  adduced  might  be  some 
what  more  rigorous,  the  Statics  of  Monge  is  a  model  of  logic, 
clearness,  and  simplicity. 

At  an  epoch  when  the  public  distress  called  into  the  higher 
ranks  all  useful  and  courageous  talent  to  the  succor  of  the 


10  BIOGRAPHICAL   NOTICE. 

country,  which  was  threatened  with  an  invasion,  Monge  was 
appointed  Minister  of  the  Navy.  He  did  every  thing  to  retain 
in  France,  men  eminent  for  their  merit  or  bravery ;  and  de 
scended  even  to  petition,  to  obtain  the  continuation  of  the 
services  of  Borda,  in  which  he  had  the  good  fortune  to  suc 
ceed.  He  was  one  of  the  most  active  men  in  scientific  labors  for 
the  safety  of  the  state.  To  him  is  due  the  construction  of  the 
new  grinding  apparatus  in  the  powder  works  of  Grenelle,  and 
the  boring  machines  on  the  boats  of  the  Seine.  He  spent 
the  daytime  in  giving  instructions  and  directions  in  the  work 
shops,  and  the  nights  in  preparing  his  Treatise  on  the  Art  of 
making  Cannons;  a  work  intended  as  a  manual  for  the  di 
rectors  of  foundries,  and  for  artizans. 

It  was  in  his  course  of  lectures  at  the  Normal  School  that 
he  delivered,  for  the  first  time,  his  Lessons  on  Descriptive 
Geometry,  the  secrets  of  which  he  had  not  been  able  sooner 
to  reveal.  Another  institution,  which  preceded  the  Normal 
School  in  the  order  of  conception,  but  which,  being  longer  in 
maturing  by  its  authors,  followed  it  in  the  order  of  execution, 
realized  in  part  the  hopes  that  had  been  conceived  in  vain  of 
founding  the  first  Universal  School  opened  in  France.  Monge, 
by  uniting  the  result  of  long  experience  at  Mezieres  with  his 
original  and  profound  views,  arranged  the  plan  of  studies, 
indicated  their  connection,  and  proposed  practicable  means  of 
execution.  Out  of  four  hundred  pupils,  who  had  first  entered 
the  Polytechnic  School,  fifty  of  the  best  scholars  were  taken 
to  form  a  preparatory  school ;  and  it  was  Monge  almost  alone, 
who  trained  them.  He  remained  the  whole  day  amongst  them, 
giving  them  by  turns  lessons  in  Geometry  and  Analysis ;  ex 
horting,  encouraging,  and  inflaming  them,  by  that  ardor,  be 
nevolence,  and  impetuosity  of  genius,  which  caused  him  to 
exhibit  to  his  pupils  the  truths  of  science  with  irresistible 
force  and  charm.  In  the  evening,  Monge  commenced  his 
labors  anew :  he  wrote  the  pages  of  Analysis  which  were  to 


BIOGRAPHICAL    NOTICE.  11 

serve  for  the  text  of  his  next  lessons ;  and  on  the  following 
day  he  was  with  his  pupils  at  the  first  moment  of  their  assem 
bling. 

The  amiability  of  Monge  was  neither  the  premeditation  of 
the  sage,  nor  even  the  effect  of  education  :  it  was  an  unaffected 
gentleness  of  disposition,  which  he  owed  to  his  happy  organi 
zation.  He  was  born  to  love  and  admire.  He  was  excessive 
both  in  his  admiration  and  his  love ;  hence  he  did  not  always, 
perhaps,  remain  within  the  limits  where  passionless  and  cold 

reason  would  have  restrained  him As  he  was  the 

father  of  the  pupils  in  the  bosom  of  the  school,  he  was  like 
wise  the  friend  of  the  soldier  in  camp. 

While  travelling  through  Italy,  collecting  the  statues  and 
paintings  ceded  to  France  by  treaty,  Monge  had  been  struck 
by  the  strange  contrast  which  the  monuments  of  the  Greeks 
presented  to  those  of  the  Egyptians,  transported  to  the 
banks  of  the  Tiber  by  Augustus  and  his  successors.  The 
comparative  character  of  the  antique  monuments  became  the 
subject  of  frequent  conversation  between  the  conqueror  of 
Italy  and  the  commissioner,  who  collected  for  his  country 
the  most  beautiful  of  the  fruits  of  victory.  Monge  con 
ceived  the  idea  of  extending  the  domain  of  history  beyond  the 
fabulous  ages  of  Greece ;  of  ascertaining,  with  the  certainty 
of  geometry,  what  the  works  of  the  ancient  sages  of  the  East 
were  ;  of  recovering,  by  contemplating  their  monuments,  that 
which  had  been — the  processes  of  their  arts,  the  customs  of 
their  public  life,  the  order  and  majesty  of  their  feasts  and 
ceremonies. 

Charged  by  the  General-in-chief,  Napoleon,  with  bringing 
the  treaty  of  Campo  Formio  to  the  Directory,  lie  was,  shortly 
afterwards,  in  the  first  rank  of  those  who  composed  the  Com 
mission  of  Science  and  Art,  which  accompanied  the  Egyptian 
expedition.  He  was  the  first  President  of  the  Institute  of 
Egypt,  formed  upon  the  model  of  the  Institute  of  France.  He 


12  BIOGRAPHICAL   NOTICE. 

visited  the  Pyramids  twice,  saw  the  obelisk  and  the  great  walls 
of  Heliopolis,  and  studied  the  remains  of  antiquity  scattered 
around  Cairo  and  Alexandria.  During  a  wearisome  march  in 
the  interior  of  the  desert,  he  discovered  (as  he  supposed)  the 
cause  of  that  wonderful  phenomenon  known  as  the  mirage. 
At  the  time  of  the  revolt  at  Cairo,  there  were  only  a  few  de 
tachments  of  troops  in  the  city,  and  the  palace  of  the  Institute 
was  guarded  only  by  the  scientific  corps.  It  was  proposed 
that  they  should  make  their  way,  sword  in  hand,  to  the  head 
quarters  ;  but  Monge  and  Berthollet,  considering  that  the  palace 
contained  books,  manuscripts,  plans,  and  antiquities,  the  fruits 
of  the  expedition,  maintained  that  the  preservation  of  this  pre 
cious  deposit  was  the  first  duty  of  the  scientific  corps;  and 
they  decided  to  die,  if  necessary,  in  defending  this  treasure, 
rather  than  to  .desert  it. 

Monge  presided  over  the  Commission  of  Science  and  Art  of 
Egypt;  and  by  his  counsels,  contributed  greatly  to  the  judi 
cious  conception  of  the  plan,  its  arrangement,  the  proportion 
of  the  principal  parts,  and  the  means  of  improving  the  arts  of 
execution. 

He  had  an  inimitable  manner  of  expounding  the  most  ab 
stract  truths,  and  of  rendering  them  clear  by  the  language  of 

action It  was,  however,  only  by  combating  nature 

that  he  became  so  excellent  a  professor :  for  he  spoke  with 
difficulty,  indeed  almost  stammered;  his  utterance,  causing 
him  to  drawl  some  syllables  and  utter  others  with  too  great 
rapidity.  His  physiognomy,  habitually  calm,  presented  an 
aspect  of  meditation;  but  when  he  spoke,  he  appeared  sud 
denly  as  though  a  different  man  :  a  new  fire  instantly  lighted 
up  his  eyes,  his  features  became  animated,  and  his  figure  in 
spired 

Enfeebled  by  age,  Monge  became  the  victim  of  an  imagina 
tion,  which,  according  as  the  times  were  adverse  or  propitious, 
carried  him  beyond  well  founded  fears  or  hopes 


BIOGRAPHICAL   NOTICE.  13 

The  regulations  of  the  service  did  not  permit  generous  youth, 
at  his  funeral,  to  deposite  the  palm  of  gratitude  and  regret  upon 
the  tomb  of  their  first  benefactor;  but,  with  the  early  dawn 
which  followed  the  day  of  his  obsequies,  the  pupils  silently 
wended  their  way  to  his  place  of  sepulture,  and  deposited  there 
a  branch  of  oak,  to  which  they  suspended  a  laurel  crown. 
Twenty-three  of  the  former  pupils  of  the  Polytechnic  School, 
all  residents  of  the  city  of  Douai,  united  spontaneously,  and 
decided  to  write  in  common  to  M.  Berthollet,  begging  him  to 
superintend  the  erection  of  a  monument  to  be  raised  at  the 
expense  of  the  old  pupils  of  the  Polytechnic  School,  in  honor 
of  Gaspard  Monge. 


TABLE    OF    CONTENTS. 


PAGK 

BIOGRAPHICAL  NOTICE  OF  MONGE 7 

DEFINITIONS 15 

CHAP.  I. — ON  THE  COMPOSITION  AND  DECOMPOSITION  OF  FORCES  17 

II. — ON  MOMENTS 50 

III. — ON  CENTRES  OF  GRAVITY 87 

IV. — ON  THE  EQUILIBRIUM  OF  MACHINES 118 

ART.  I. — On  the  Equilibrium  of  Forces  which 
act  upon  each  other  by  means  of 
Cords 120 

ART.  II. — On  the  Equilibrium  of  the  Lever 130 

Of  Pulleys 141 

Of  the  Wheel  and  Axle 153 

Of  Cog  Wheels ICG 

Of  the  Jack  Screw 1G8 

ART.  III. — On   the  Equilibrium  of  the  Inclined 

Plane 169 

Of  the  Screw 185 

Of  the  Wedge 193 

NOTE,  CONTAINING  A  DEMONSTRATION  OF  THE  PARALLELOGRAM 

OF  FORCES,  BY  M.  CAUCHY 203 

14 


AN 


ELEMENTARY   TREATISE 


ON 


STATICS. 


DEFINITIONS. 

EVERY  thing  that  is  capable  of  affecting  our  senses  is 
called  a  body,  or  material  substance. 

Bodies  are  divided  into  solids  and  fluids.  A  body  is 
solid  when  the  molecules  which  compose  it  are  cohesive, 
and  cannot  be  displaced  among  each  other  without 
effort:  the  metals,  stones,  wood,  &c.,  are  of  this  num 
ber.  It  is  fluid  when,  on  the  contrary,  all  its  molecules 
can  be  separated  with  the  greatest  ease ;  such  are  water, 
air,  &c. 

All  bodies  are  moveable,  that  is,  they  may  be  trans 
ported  from  one  place  to  another.  A  body  is  said  to  bo 
at  rest,  when  all  the  parts  which  compose  it  remain  each 
in  the  same  place ;  and  it  is  said  to  be  in  motion,  when 
it  changes  place,  or  when  the  parts  of  which  it  is  com 
posed  pass  from  one  place  to  another. 

A  body  at  rest  cannot  enter  into  motion,  and  when 
in  motion,  cannot  change  the  manner  in  which  it  moves, 

15 


16  STATICS. 

without  the  action  of  some  cause,  to  which  has  been 
given  in  general  the  name  of  force  or  power. 

There  are  to  be  considered  in  a  force,  1st,  its  intm- 
sity,  that  is  to  say,  the  effort  which  it  makes  to  move 
the  body,  or  the  particle  of  the  body  to  which  it  is  ap 
plied  ;  2d,  its  direction,  or  the  straight  line  in  which  it 
tends  to  move  the  point  of  the  body  upon  which  it  acts. 

When  several  forces  are  applied  to  the  same  body, 
two  cases  may  happen :  either  these  forces  counter 
balance  and  mutually  destroy  each  other,  when  they 
are  said  to  be  in  equilibrium  ;  or,  by  reason  of  the  action 
of  all  these  forces,  the  body  enters  into  motion. 

Hence,  the  term  Mechanics  is  given  to  that  science 
whose  object  it  is  to  find  the  effect  which  the  applica 
tion  of  determined  forces  must  produce  upon  a  body. 
This  science  is  divided  into  two  parts :  the  first  con 
siders  the  relations  which  the  forces  should  have,  in  in 
tensity  and  direction,  so  as  to  be  in  equilibrium,  and  is 
called  Statics;  the  second,  to  which  the  name  Dynamics 
has  been  given,  determines  the  manner  in  which  the  body 
moves,  when  these  forces  do  not  entirely  destroy  each 
other. 

Each  of  these  parts  is  again  divided  into  two  others, 
according  as  the  body,  to  which  the  forces  are  supposed 
to  be  applied,  is  solid  or  fluid.  The  part  of  Statics 
which  treats  of  the  equilibrium  of  forces  applied  to 
solid  bodies,  is  named  simply  Statics,  or  Statics  proper; 
and  Hydrostatics  is  that  which  has  for  its  object  the 
equilibrium  of  forces  applied  to  the  different  molecules 
of  a  fluid  body. 

In  this  Treatise  we  will  consider  only  the  first  of 
these  two  parts,  that  is,  Statics  proper. 


IT 


CHAPTER  FIRST. 

OF   THE   COMPOSITION  AND  DECOMPOSITION   OF   FORCES. 

1.  When  a  force  p,  applied  to  a  determined  point  c 
of  a  solid  body  AB,  draws  or  pushes  this  body  in  any 
direction  CF,  we  may  consider  this  force  as  though  it 
were  immediately  applied  to  any  other  point  D  of  the 
body,  taken  upon  the  direction  of  this  force. 

For  all  the  points  of  the  body,  which  are 
in  the  straight  line  CF,  can  neither  approach 
towards,  nor  remove  from,  each  other;  and 
none  of  them  can  move  along  this  line  with 
out  causing  all  the  others  to  move  in  the 
same  manner  as  though  the  force  were  im 
mediately  applied  to  them. 

We  are  likewise  permitted  to  consider  the  force  p  as 
though  it  were  applied  to  any  other  point  G,  taken  beyond 
the  body,  upon  its  direction,  provided  this  point  is  in 
variably  attached  to  the  body. 

2.  Hence  it  follows  that  if,  upon  the  direction  of  the 
force  P,  there  is  found  a  fixed  point  D  within  the  body, 
or  an  immovable  obstacle  beyond  it,  provided   in  the 
latter  case  the  obstacle  is  invariably  attached  to  the 
body,  the  force  will  be  destroyed,  and  the  body  will  re 
main  at  rest ;  for  this  force  may  be  regarded  as  imme 
diately  applied  to  the  fixed  point,  and  its  effect  will  be 
destroyed  by  the  resistance  of  this  point. 

2* 


c 
b 


18  STATICS. 

xfy-  2.          3.  Reciprocally,  if  the  force  P,  applied  to  the 
body  AB,  is  destroyed  by  the  resistance  of  a 
single  fixed  point,  this  point  is  found  upon 
the  direction  of  the  force  :  for  this  point  can 
r.  B  destroy  the  effect  of  the  force  only  by  op 
posing  the  motion  of  the  point  of  application 
o      c ;  and  it  cannot  prevent  this  motion,  unless 
it  is  upon  the  straight  line  which  the  force  tends  to 
make  the  point  of  application  traverse. 

4.  A  point  cannot  move  in  several  directions  at  the 
same  time. 

5.  When  several  forces,  differently  directed,  are  ap 
plied  at  the  same  time  to  the  same  point,  either  this 
point  will  remain  at  rest,  or  it  will  move  in  a  single  di 
rection,  and  consequently  in  the  same  manner  as  though 
it  were  pushed  or  drawn  by  a  single  force  along  this  di 
rection  and  capable  of  the  same  effect. 

6.  Thus,  whatever  may  be  the  number  and  directions 
of  the  forces  applied  at  the  same  time  to  the  same 
point,  there  always  exists  a  single  force  which  can  move 
it,  or  tends  to  move  it  in  the  same  manner  as  all  these 
forces  together ;  this  single  force  is  named  the  resultant 
of  the  former  ;  and  they,  with  reference  to  the  resultant, 
are  named  the  component  forces. 

The  operation,  by  which  we  seek  the  resultant  of 
several  given  component  forces,  is  named  the  composi 
tion  of  forces ;  and  that  by  which  we  find  the  compo 
nents,  when  the  resultant  is  known,  is  named  the  de 
composition  of  forces. 

7.  Two  forces  are  equal,  when,  being  applied  to  the 
same  point  and  directly  opposed,  they  destroy  each  other 
and  produce  an  equilibrium. 


COMPOSITION    OF   FORCES.  19 

Reciprocally,  when  two  forces  are  in  equilibrium,  they 
are  equal  and  directly  opposed. 

8.  For  if  several  forces,  differently  directed,  be  ap 
plied  to   the  same  point,   it  is  necessary,   in   order  to 
put  them  in  equilibrium,  or  to  destroy  the  effect  of  their 
resultant,  that  a  single   force,  equal  to  this  resultant, 
should  be  applied  to  this  point  and  directly  opposed  to 
it,  or  that  several  forces  be  applied,  the  resultant  of 
which  is  equal  and  directly  opposed  to  the  resultant  of 
the  former. 

9.  Reciprocally,  when  several  forces,  differently  di 
rected  and  applied  to  the  same  point,  are  in  equilibrium, 
their  resultant  is  zero ;  or  what  is  the  same  thing,  any 
one  of  these  forces  is  equal  and  directly  opposed  to  the 
resultant  of  all  the  others ;  or,  lastly,  the  resultant  of 
any  number  of  these  forces  is  equal  and  directly  opposed 
to  the  resultant  of  all  the  others. 

10.  The  resultant  of  two  forces,  applied  to  the  same 
point,  is  in  the  plane  determined  by  the  directions  of  these 
forces ;    and  it  is  necessarily  included  in  the  angle 
formed  by  these  two  forces. 

If  the  resultant  were  not  in  the  plane  of  the  two 
forces,  there  would  be  no  reason  why  it  should  be  above 
the  plane  rather  than  below:  it  cannot  be  at  the  same 
time  in  two  different  positions  ;  hence  it  is  in  the  plane 
of  the  two  forces ;  moreover,  it  is  included  in  the  angle 
of  the  lines  along  which  these  forces  are  directed ;  for 
there  is  no  force  tending  to  move  the  point  into  the 
space  adjacent  to  this  angle ;  hence  it  will  remain  in 
the  angle  itself. 

11.  A  force  is  the  multiple  of  another  force  when 


20  STATICS. 

it  is  formed  ly  the  union  of  several  forces  equal  to  the 
latter. 

Thus,  if  we  apply  to  the  same  point  and  in  the  same 
direction  several  forces  equal  to  each  other,  and  if  we 
take  any  one  of  these  forces  as  unity,  the  multiple  force 
will  be  expressed  by  a  number  equal  to  that  of  the 
added  forces. 

As  it  is  always  possible  to  compare  numbers  with 
lines,  we  may  represent  a  force  by  a  straight  line  taken 
upon  its  direction,  and  its  multiple  force  by  another 
straight  line,  a  multiple  of  the  former ;  in  Statics,  fre 
quent  use  is  made  of  this  geometrical  construction. 

12.  Three  equal  forces,  which  are  applied  to  the 
same  point9  and  which  divide  the  circumference,  of  which 
this  point  is  the  centre,  into  three  equal  parts,  are  ne 
cessarily  in  equilibrium. 

The  point  to  which  these  three  forces  are  applied  can 
move  only  in  a  single  direction ;  but  the  line  along 
which  it  would  move,  could  be  placed  in  three  or  six 
manners  perfectly  similar  with  reference  to  the  three 
forces ;  now  there  is  no  reason  that  the  motion  of  the 
point  should  take  place  in  one  direction  rather  than  in 
another ;  hence  it  will  remain  at  rest. 

The  application  of  these  three  equal  forces  to  the 
same  point  proves,  that  there  is  evidently  for  this  case 
a  resultant  of  these  forces,  since  any  one  of  the  three 
forces  makes  an  equilibrium  with  the  two  others,  and  it 
is  evident  that  the  direction  of  one  of  these  forces  di 
vides  the  angle  formed  by  the  two  others,  into  two  equal 
parts ;  it  is  not  less  evident  that  the  resultant  of  any 
two  equal  forces  meeting  in  a  point,  divides  the  angle 
formed  by  these  two  forces  into  two  equal  parts. 


COMPOSITION    OF    FORCES.  21 

13.  If  several  forces  are  applied  to  the  same  point 
along  the  same  line  and  act  in  the  same  direction,  it  fol- 
loivs,  from  proposition  No.  11,  that  their  resultant  is  a 
single  force,  equal  to  their  sum,  acting  along  the  same 
line,  and  in  the  same  direction. 

Hence,  to  make  an  equilibrium  of  all  these  forces,  it 
is  necessary  to  apply  to  the  same  point  and  in  the  oppo 
site  direction,  a  force  equal  to  their  sum ;  for  this  force 
will  be  equal  and  directly  opposed  to  their  resultant. 

14.  From  this  it  follows :  1st.  If  two  unequal  forces 
are  applied  to  the  same  point  in  opposite  directions, 
their  resultant  is  in  the  direction  of  the  greater,  and  is 
equal  to  their  difference :  for  the  greater  of  these  two 
forces   may  be   regarded  as  composed  of   two   others 
having  the  same  direction ;  one  of  which  is  equal  to  the 
smaller  force,  and  the  other  equal  to  the  difference  : 
now  the  first  of  these  two  latter  forces  is  destroyed  by 
the  smaller  (7) ;  then  to  move  the  point,  there  remains 
only  the  difference,  which  is  in  the   direction   of  the 
greater. 

2d.  If  any  number  of  forces  are  applied  to  the  same 
point,  some  being  placed  in  one  direction,  and  the  others 
directly  opposed,  after  having  taken  the  sum  of  all 
those  acting  in  one  of  the  two  directions,  and  the  sum 
of  all  those  acting  in  the  contrary  direction,  the  re 
sultant  of  all  these  forces  is  equal  to  the  difference  of 
the  two  sums  (12),  and  is  in  the  direction  of  the  greater. 

Hence,  to  produce  an  equilibrium  of  all  these  forces, 
it  is  necessary  to  apply  to  the  same  point,  and  along 
the  direction  of  the  smaller  of  the  two  sums,  a  force 
equal  to  the  difference  of  these  sums ;  for  this  force 
will  be  equal  and  directly  opposed  to  their  resultant. 


22 


STATICS. 


THEOREM. 

15.  If  at  the  extremities  of  an  inflexible  straight 
line  AB,  two  equal  forces  P,  Q,  be  applied  along  the  lines 
AP,  BQ,  parallel  to  each  oilier,  and  which  act  in  the 
same  direction  : 

1st.  The  direction  of  the  resultant  R  of  these  tivo 
forces  is  parallel  to  the  straight  lines  AP,  BQ,  and  passes 
through  the  middle  of  AB  ; 

2d.  This  resultant  is  equal  to  the  sum  P-fQ  of  the 
two  forces. 

DEMONSTRATION. — Let  another  in 
flexible  straight  line  DE  be  drawn 
perpendicular  to  the  directions  of  the 
two  forces  P  and  Q,  and  invariably 
connected  with  the  straight  line  AB : 
having  prolonged  the  directions  of 
the  two  forces,  we  may  suppose  that 
these  two  forces  act  (1)  at  the  points 
D  and  E  ;  moreover,  we  may  apply  to  the  same  points, 
the  forces  p,  p',  and  q,  q'9  equal  to  the  forces  P  and  Q, 
so  that  the  three  equal  forces  P,  p,  p',  meeting  at  the 
point  D,  will  divide  the  circumference,  which  has  its  cen 
tre  at  the  point  D,  into  three  equal  parts,  and  so  that 
the  three  forces  Q,  q,  q,'  equal  to  each  other  and  to  the 
first,  will  divide  in  like  manner  the  circumference  which 
has  its  centre  at  the  point  E,  into  three  equal  parts. 

We  have  seen  (12)  that  the  force  P  is  equal  and  op 
posed  to  the  resultant  of  the  couple  p,  pr ;  that  the 


COMPOSITION   OF   FORCES.  23 

force  Q  is  equal  arid  opposed  to  the  resultant  of  the 
couple  g,  q' ;  hence  the  forces  P  and  Q  have  a  resultant 
equal  and  opposed  to  that  of  the  system  of  the  two 
couples  g,  qr  and  p,  p' ;  the  two  forces  pf  q\  applied  to 
the  point  F  of  their  direction,  have  for  resultant  a  third 
force  equal  to  each  of  the  first  two,  and  acting  in  the 
direction  CF ;  likewise  the  two  forces  p  and  q,  applied 
to  the  point  G  in  the  line  of  their  direction,  have  for 
resultant  a  third  force  equal  to  each  of  them,  and  di 
rected  along  the  line  GO  ;  hence,  the  resultant  of  the 
four  forces  p,  pf,  q,  qr  is  a  single  force  equal  to  two  of 
the  others,  and  acting  along  the  line  GCF,  which  divides 
the  line  DE  into  two  equal  parts ;  hence  the  resultant  of 
the  two  equal  forces  P  and  Q  is  equal  to  the  sum  P-f  Q, 
and  divides  the  line  DE,  or  the  line  AB,  to  which  these 
forces  are  applied,  into  two  equal  parts.  (See  a  second 
demonstration  of  this  theorem,  No.  19.) 


COROLLARY  I. 

16.  Hence,  to  bring  the  two  forces  P  and  Q  into  an 
equilibrium,  it  is  necessary  to  apply  to  the  middle  K  of 
the  straight  line  AB  a  third  force  equal  to  their  sum, 
which  will  act  in  a  contrary  direction,  and  which   di 
rection  will  be  parallel  to  the  two  lines  AP,  BQ  ;  for  this 
third  force  will  be  equal  and  directly  opposed  to  their 
resultant. 

COROLLARY  II. 

17.  If  an  inflexible  straight  line  be  divided  into  any 
number  of  equal  parts,  and  we  apply  to  all  the  points 


24  STATICS. 

of  division,  equal  and  parallel  forces,  the  resultant  of 
all  these  forces  will  pass  through  the  middle  of  the  line, 
in  a  direction  parallel  to  that  of  the  forces,  and  will  be 
equal  to  their  sum. 

For,  all  the  partial  resultants  of  these  forces,  con 
sidered  two  and  two,  and  taken  at  equal  distances  from 
the  middle  point  of  the  line,  will  pass  through  this  point 
(15),  in  the  direction  of  the  forces,  and  each  of  them 
will  be  equal  to  the  sum  of  the  two  forces  which  com 
pose  it ;  hence  (13)  the  general  resultant  will  also  pass 
through  the  middle  of  the  line,  in  the  same  direction, 
and  will  be  equal  to  the  sum  of  all  the  partial  result 
ants,  that  is  to  say,  to  the  sum  of'  all  the  component 
forces. 


THEOREM. 

18.  If  at  the  extremities  of  an  inflexible  straight 
line,  two  unequal  forces  P  and  Q  be  applied,  tvhose  lines 
of  direction  AP,  BQ,  are  parallel  to  each  other,  and 
act  in  the  same  direction  : 

1st.  The  resultant  R  of  these  two  forces  is  equal  to 
their  sum,  and  its  direction  is  parallel  to  that  of  these 
forces  ; 

2d.  The  point  of  application  c  of  the  resultant  di 
vides  the  line  AB  into  two  parts  reciprocally  propor 
tional  to  the  two  forces,  so  that  we  have 

p  :  Q  : :  BC  :  AC. 


COMPOSITION   OF   FORCES.  25 

DEMONSTRATION. — Suppose,  in 
the  first  place,  the  two  forces  P  and 
Q  are  commensurable ;  divide  the 
line  AB  into  two  parts  AD,  DB,  pro 
portional  to  the  forces  P,  Q ;  laying 
off  the  line  AD  from  A  to  E,  and 
the  line  BD  from  B  to  P,  the  line  EF  will  be  double  AB  ; 
and  since 

p  :  Q  : :  AD  :  DB, 
we  shall  have  also, 

p  :  Q  : :  ED  :  DF. 

Dividing  the  lines  AD,  DB  into  as  many  parts  as  there 
are  units  in  the  forces  P  and  Q,  and  repeating  this  di 
vision  upon  the  lines  AE,  BF,  the  whole  line  EF  will  be 
divided  into  twice  as  many  equal  parts  as  there  are 
units  in  the  sum  of  the  two  forces  P  and  Q ;  now,  the 
middle  of  each  of  these  divisions  may  be  considered  as 
the  points  of  application  of  forces  equal  to  each  other 

p_f_  Q 

and  each  equal  to  -7, —  (IT),   m  being  the  number   of 

units  of  the  sum  P+Q;  hence  the  resultant  of  all  these 
forces  will  be  also  the  resultant  of  the  two  forces  P  and 
Q ;  but  the  resultant  of  equal  forces,  distributed  on  each 
side  of  the  middle  of  equal  divisions  of  the  same  line, 
is  equal  to  the  sum  of  these  forces,  passes  through  the 
middle  of  this  line,  and  acts  in  the  same  direction  as 
the  components :  hence  the  resultant  of  the  two  forces 
p  and  Q  is  equal  to  the  sum  P+Q,  passes  through  the 
middle  point  c  of  the  line  EF,  and  acts  in  the  direction 
of  the  forces  P  and  Q ;  moreover,  from  the  construction 
of  Figure  4,  CF= JEF=AB  ;  then  subtracting  the  com 
mon  part  CB,  we  have  AC=BF=BD.  For  the  same  reason 

3 


26 


STATICS. 


EC=JEF=AB  ;  hence  CB=AE=AD  ;  hence  the  middle 
point  c  of  EF  is  such  that  we  have  P  :  Q  : :  CB  :  CA  ; 
hence  this  point  of  application  divides  the  line  AB  into 
two  parts  reciprocally  proportional  to  the  forces. 

19.  Let  us  suppose  now  that  the  two  forces  P  and  Q 
are  incommensurable.  Apply  to  the  points  A  and  B, 
and  along  the  line  AB,  two  equal  and  opposed  forces  s,  s' ; 

the  resultant  of  the  two 
forces  P  and  Q  will  be  the 
same  as  the  resultant  of 
the  four  forces  P,  s,  Q,  s' ; 
the  couple  P,  s  has  for  re 
sultant  a  line  Ai  included  in 
the  angle  SAP  ;  the  couple 
Q,  s7  has  for  resultant  a  line 
directed  along  BI  included  in  the  angle  QBS'  ;  supposing 
these  two  resultants  to  be  applied  to  the  point  of  inter 
section  I  of  their  directions,  and  drawing  through  the 
point  I  a  line  GH  parallel  to  AB,  they  will  each  be  de 
composed  into  two  forces  along  the  directions  IH  and 
GH  ;  the  forces  in  the  direction  GH  being  equal  to  s  and 
s'  and  opposed,  they  will  destroy  each  other  ;  the  forces 
in  the  direction  IR  combine  together  and  are  equal  to 
P+Q;  hence,  whether  the  two  parallel  and  unequal 
forces  P  and  Q  be  commensurable  or  incommensurable 
their  resultant  is  parallel  to  their  direction  and  equal  to 
their  sum.* 


*  This  demonstration  is  also  true  when  the  two  forces  P  and  Q 
are  equal  to  eacn  other ;  consequently  if  is  applicable  to  the  theorem 
of  No,  15. 


COMPOSITION    OF   FORCES.  27 

Fi9-  4-  20.  The  resultant  R  of  the  two 

forces  P  and  Q,  supposed  to  be  in 
commensurable,  passes  through  a 
point  c  in  such  a  manner  that  we 
have 

p  :  Q  : :  CB  :.  CA. 

For  if  it  passed  through  another  point  H,  situated  be 
tween  A  and  c,  we  should  find  a  force  Q'  applied  in  the 
direction  Q,  which  would  be  such  that  the  resultant  of 
the  commensurable  forces  P  and  Q'  would  pass  through 
a  point  K  situated  between  the  points  c  and  H ;  this 
force  Q'  would  be  the  fourth  term  of  the  proportion 

KB  :  KA  : :  P  :  Q', 
the  lines  KB,  KA,  having  for  unit  of  measure  a  line 

smaller  than  CH  ;  but  we  have  — >  —  :  hence  the  ra- 

KA       CA 
P    .  P 

tio  —  is  greater  than  the  ratio  -.     Hence  Q'  is  smaller 

than  Q :  consequently,  the  resultant  of  the  two  forces 
P  and  Q  will  necessarily  pass  between  the  points  K  and 
Q'  ;  hence  it  is  absurd  to  suppose  that  it  passes  through 
the  point  H  situated  beyond  the  point  K.  It  might  be 
demonstrated  in  the  same  manner  that  it  cannot  pass 
through  a  point  H'  situated  between  c  and  B ;  hence  it 
passes  necessarily  through  the  point  c. 

COROLLARY  I. 

21.  Hence,  in  order  that  equilibrium  may  ensue  be 
tween  the  two  forces  P$  Q,  it  is  necessary  to  divide  the 
line  AB  at  the  point  c  into  two  parts  reciprocally  pro 
portional  to  these  two  forces,  and  apply  to  the  point  c 


28  STATICS. 

a  third  force,  equal  to  the  sum  P+Q,  parallel  to  them 
and  acting  in  the  opposite  direction. 

Remark. 

22.  If  the  ratio  of  the  forces  P,  Q,  and  the  length  of 
the  line  AB,  -were  given  in  numbers,  and  it  were  desired 
to  find  the  distances  of  the  point  c  from  the  points  A, 
B,  the  proportion 

p  :  Q  : :  EC  :  AC 

could  not  be  employed  directly,  because  in  this  propor 
tion  we  would  know  only  the  first  two  terms ;  but  it  is 
easy  to  deduce  from  it  the  following, 

P+Q  :  Q  : :  BC+AC  :  AC, 
which,  since  BC+AC  is  equal  to  AB,  becomes 

P+Q  :  Q  : :  AB  :  AC, 
in  which  the  first  three  terms  are  known. 

The  distance  BC  can  be  found  by  the  proportion 

P  +  Q  :  P  : :  AB  :  BC, 
which  is  likewise  deduced  from  the  first. 

COROLLARY  II. 

23.  When  a  single  force  R  is  applied  to  a  point  c  of 
an  inflexible  straight  line,  it  may  always  be  decomposed 
into  two  others  P,  Q,  which,  being  applied  to  the  two 
points  A,  B,  given  upon  the  same  line,  and  being  di 
rected  parallel  to  RC,  produce  the  same  efiect ;  and  the 
intensity  of  the  two  forces  is  found  by  dividing  the 
force  R  into  two  parts  reciprocally  proportional  to  the 
lines  AC,  CB,  by  means  of  the  two  following  proportions  : 

AB  :  BC  : :  R  :  P, 

AB  :  AC  :  :  R  :  Q, 

in  each  of  which  the  first  three  terms  are  known ;  for 


COMPOSITION    OF   FORCES.  29 

the  resultant  of  the  two  forces  p,  Q,  has  the  same  in 
tensity,  is  parallel  to,  and  acts  in  the  same  direction  as, 
the  force  R. 


COROLLARY  III. 

24.  Since  Fig.  6   is  the  same  in  all  respects  as  the 
preceding,  if  we  apply  to  the  point  c  of  the  line  AB  a 

19 '^  '  force  s,  equal  and  directly  opposed  to 

the  resultant  of  the  two  forces  P,  Q,  in 
such  manner  that  we  have 
S=R=P+Q  (19), 

the  three  forces  p,  Q,  s,  will  be  in  equi 
librium,  and  each  of  the  two  forces  P,  Q, 
may  be  regarded  as  equal  and  directly 
opposed  to  the  resultant  of  the  two  others.  Hence  the 
resultant  of  the  two  forces  s,  Q,  which  are  parallel,  and 
act  in  contrary  directions,  is  a  force  p  equal  and  directly 
opposed  to  the  force  P.  Now,  the  force  P  is  equal  to 
the  difference  of  the  forces  s,  Q,  and  acts  in  the  con 
trary  direction  to  the  greater,  s,  of  these  two  forces : 
hence,  1st,  the  resultant  p  or  — p  of  the  two  forces  S,  Q, 
is  equal  to  their  difference  s — Q,  and  acts  in  the  direction 
of  the  greater  and  parallel  to  these  two  forces. 
Moreover,  we  have  P+Q,  or  s  :  Q  : :  AB  :  AC  (20). 
Hence,  2d,  the  distances  of  the  point  of  application  A 
of  this  resultant  from  the  two  points  c,  B,  are  recipro 
cally  proportional  to  the  forces  S,  Q. 

Remark. 

25.  If  the  ratios  of  the  two  forces  s,  Q,  and  the 
length  of  the  line  EC,  were  given    in  numbers,  and 


30  STATICS. 

it  were  desired  to  find  the  distances  of  the  point  A 
from  the  points  B,  c,  the  preceding  proportion  could  not 
be  employed  directly,  because  only  the  first  two  terms 
of  this  proportion  would  be  known ;  but  it  is  easy  to 
deduce  the  following  from  it : 

s— Q  :  Q  :  :  AB— AC  or  BC  :  AC, 
in  which  the  three  first  terms  are  known. 

The  distance  AB  can  be  found  by  another  proportion, 

s— Q  :  S  : :  AB— AC  or  BC  :  AB, 
which  likewise  is  deduced  from  the  first. 

COROLLARY  IV. 

26.  If  the  two  forces  s,  Q,  which  are  parallel,  and 
act  in  contrary  directions,  be  equal  to  each  other,  1st, 
their  resultant  P,  which  is  equal  to  S — Q  (24),  becomes 
zero;  2d,  in  the  proportion  s— Q  :  Q  : :  BC  :  AC,  the 
second  term  being  infinitely  great  compared  with  the 
first  which  is  zero,  the  fourth  term  is  also  infinitely  great 
compared  with  the  third.  Hence  the  point  of  applica 
tion  A  of  the  resultant  P  is  at  an  infinite  distance  from 
the  point  c ;  hence,  to  produce  an  equilibrium  between 
the  two  forces  s,  Q,  it  would  be  necessary  to  apply  to 
the  inflexible  straight  line  a  zero  force,  whose  line  of  di 
rection  should  be  at  an  infinite  distance ;  which  is  not 
absurd,  but  which  cannot  be  executed. 

We  perceive,  then,  that  it  is  impossible,  by  means  of 
a  single  force,  to  produce  an  equilibrium  between  two 
equal  and  parallel  forces,  acting  in  contrary  directions ; 
but  it  will  be  demonstrated  (51),  that,  by  means  of  two 
forces,  an  equilibrium  can  be  produced  between  them  in 
an  infinite  number  of  ways. 


COMPOSITION    OF    FORCES. 


31 


PROBLEM. 

27.  Any  number  of  parallel  forces  P,  Q,  R,  s,  .  .  .  . 
which  act  in  the  same  direction,  being  applied  to  the 
points  A,  B,  c,  D,  .  .  .  .  of  given  position,  and  connected 
together  in  an  invariable  manner,  to  determine  the  re 
sultant  of  all  these  forces. 

SOLUTION. — Consider 
ing  first  any  two  of  these 
forces,  such  as  P  and  Q, 
determine  their  resultant 
T  (18);  this  resultant  will 
be  equal  to  P+Q;  its  di 
rection  will  be  parallel  to 
that  of  the  forces  P,  Q, 
and  we  will  find  its  point 
of  application  E,  by  the 
following  proportion  (22) : 

P+Q  :  Q  : :  AB  :  AE. 

In  place  of  the  two  forces  P,  Q,  substitute  their  re 
sultant  T  ;  then  having  drawn  the  line  EC,  determine 
the  resultant  v  of  the  two  forces  T,  R ;  this  resultant  V 
will  be  also  that  of  the  three  forces  P,  Q,  R  ;  its  intensity 
will  be  T  +  R  or  P+Q+R;  and  find  on  EC  its  point  of  ap 
plication  F  by  the  proportion 

T+R  or  P+Q+R  :  R  : :  EC  :  EF. 

In  place  of  the  three  forces  P,  Q,  R,  let  their  re 
sultant  v  be  substituted,  and  having  drawn  the  line  FD, 
the  resultant  x  of  the  two  forces  v,  s,  will  be  found ; 
this  resultant  x  will  be  also  that  of  the  four  forces  P,  Q, 
R,  s;  its  magnitude  will  be  v+s,  or  P+Q+R+S,  and 


32  STATICS. 

its  point  of  application  G  upon  FD  will  be  found  by  the 
proportion 

V+S,  or  P+Q+R+S  :  s  :  :  FD  :  FG. 
By  continuing  in  this  manner,  the  position  of  the 
general  resultant  of  all  the  forces  will  be  found,  what 
ever  their  number  may  be  ;  and  the  intensity  of  this  re 
sultant  will  be  equal  to  the  sum  of  all  these  forces. 

COROLLARY  I. 

28.  Hence,  by  supposing  the  point  G  to  be  connected 
with  the  other  points  A,  B,  c,  D,  ....  in  an  invariable 
manner,  an  equilibrium  will  be  produced  between  all  the 
other  forces  P,  Q,  R,  s,  .  .  .  .  by  applying  to  the  point  G 
a  force  parallel  to  the  first,  which  acts  in  the  contrary 
direction,  and  which  is  equal  to  their  sum 


COROLLARY  II. 

29.  If  among  the  parallel  forces  P,  Q,  R,  S,  .  .  .  some 
should  act  in  one  direction,  and  others  in  the  contrary 
direction,  we  could  determine,  first  (27),  the  partial  re 
sultant  of  all  which  act  in  one  direction,  and  then  the 
partial  resultant  of  all  which  act  in  the  contrary  di 
rection.  Thus  all  the  forces  would  be  reduced  to  two 
others  acting  in  opposite  directions  ;  and  by  determining, 
by  the  process  of  No.  24,  the  resultant  of  these  last 
two  forces,  supposed  to  be  unequal,  we  should  have  the 
general  resultant,  and  consequently  the  force  which, 
being  applied  in  the  contrary  direction,  would  produce 
an  equilibrium  amon^  all  the  proposed  forces  ;  if  these 


COMPOSITION   OF   FORCES.  33 

forces  should  be  reduced  to  two  parallel  and  equal 
forces,  we  have  seen  (26),  that  it  is  impossible  to  pro 
duce  an  equilibrium  between  them. 

The  general  resultant  being  equal  to  the  difference  of 
the  two  partial  resultants  (24),  and  each  of  these  being 
equal  to  the  sum  of  those  which  compose  it  (27),  it  fol 
lows  that  the  general  resultant  is  equal  to  the  excess 
of  the  sum  of  the  forces  which  act  in  one  direction 
over  the  sum  of  those  which  act  in  the  opposite  di 
rection. 

COROLLARY  III. 

30.  If  the  forces  p,  Q,  R,  s,  .  .  .  without  ceasing  to 
be  parallel  and  without  changing  in  intensity,  had 
another  direction  and  became  j9,  q,  r,  s,  .  .  .  the  result 
ant  t  of  the  first  two  would  also  pass  through  the  point 
E,  and  would  be  equal  to  the  sum  p  +  q.  Likewise  the 
resultant  v  of  the  three  forces  p,  q,  r,  would  pass  through 
the  point  F,  and  would  be  equal  to  the  sum  p+q+r. 
So  also  the  resultant  x  of  the  four  forces  p,  q,  r,  s, 
would  pass  through  the  point  G,  and  would  be  equal  to 
the  sum p-\-q-\-r+s,  and  so  on.  Hence  the  general  re 
sultant  of  all  the  forces  p,  q,  r,  s,  would  pass  through 
the  same  point  as  the  resultant  of  the  first  forces 
P,  Q,  R,  S, 

Hence  it  appears  that,  when  the  intensities  and  points 
of  application  remain  the  same,  the  resultant  of  these 
forces  always  passes  through  a  certain  identical  point, 
whatever  may  be  their  direction ;  and  the  intensity  of 
this  resultant  is  always  equal  to  their  sum. 


34  STATICS. 

The  point  through  which  the  resultant  of  the  parallel 
forces  always  passes,  whatever  may  be  their  direction, 
is  named  the  centre  of  parallel  forces. 

It  is  easy  to  see  that  if  the  points  of  application  A, 
B,  c,  D,  .  .  .  of  the  parallel  forces  P,  Q,  R,  s,  .  .  .  are  in 
the  same  plane,  the  centre  of  these  forces  is  also  in  this 
plane ;  for  this  plane  contains  the  line  AB,  and  conse 
quently  the  point  E  of  this  line,  which  is  the  centre  of 
the  forces  P,  Q :  it  contains  also  the  line  EC,  and  conse 
quently  the  centre  F  of  the  forces  P,  Q,  R  ;  it  contains 
the  line  FD,  and  consequently  the  centre  G  of  the  forces 
p,  Q,  R,  s,  and  so  on. 

It  may  be  demonstrated  in  like  manner  that,  if  the 
points  of  application  are  upon  the  same  straight  line, 
the  centre  of  the  parallel  forces  is  also  upon  this  line. 

LEMMAS. 
I. 

31.  If  a  power  P  be  applied  to  the  circumference  of 
a  circle  movable  about  its  centre  A,  and 
in  a  direction  BP  tangent  to  the  circum 
ference,  this  force  tends  to  turn  the 
circle  about  its  centre,  as  though  it  were 
applied  at  any  other  point  c,  and  in  a 
direction  CQ  tangent  to  the  same  cir 
cumference. 

II. 

A  poiver  P,  applied  along  the  direction  of  a  line  on 
which  there  is  a  fixed  point,  is  destroyed  by  the  re 
sistance  of  this  point.  (Nos.  2  and  3). 


COMPOSITION   OF  FORCES. 


35 


THEOREM. 

32.  When  the  directions  of  two  forces  P,  Q,  are  con 
tained  in  the  same  plane,  and  intersect  in  the  same 
point  A,  if  we  lay  off  in  these  directions  the  lines  AB, 
AC,  proportional  to  these  forces,  so  that  we  have 

p  :  Q  : :  AB  :  AC, 

and  complete  the  parallelogram  ABCD,  the  resultant  of 
these  two  forces  will  be  in  the  direction  of  the  diagonal 
AD  of  the  parallelogram. 

DEMONSTRATION. — Suppose, 
for  an  instant,  that  the  point 
D  of  the  diagonal  AD  is  an  im 
movable  obstacle ;  from  this 
point  let  fall  the  perpendicu 
lars  DE  DF  upon  the  directions 
of  the  two  forces ;  the  trian 
gles  BED,  CFD  will  be  similar, 
because  the  angles  at  B  and  c 
being  equal  to  A,  will  be  equal 
to  each  other,  and  we  shall  have 

DC  :  DB  : :  DF  :  DE. 
Now,  we  have  by  supposition 

p  :  Q  : :  AB  :  AC,  or  : :  DC  :  DB. 
Hence  we  shall  have 


p  :  Q  : :  DE  :  DE. 


36  STATICS. 

From  the  point  r>,  as  a  centre,  and  with  a  radius  DF,  de 
scribe  the  arc  FG,  terminating  in  the  prolongation  of  ED 
at  G  ;  then,  regarding  this  arc  and  the  line  EG  as  inflex 
ible  lines,  and  connected  in  an  invariable  manner  at  the 
point  A,  let  us  conceive  the  force  P  to  be  applied  at  the 
point  E  along  its  direction,  and  a  force  M,  equal  to  the 
force  Q,  to  be  applied  to  the  point  G,  in  a  direction 
parallel  to  AP,  and  consequently  tangent  to  the  arc 
FG.  This  being  granted,  since  M=Q  and  DF=DG,  we  will 
have 

p  :  M  : :  DG  :  DE. 

Hence  (18)  the  resultant  of  the  two  parallel  forces 
p,  M,  will  pass  through  the  fixed  point  D,  and  will 
be  destroyed  by  the  resistance  of  this  point ;  hence 
these  two  forces  will  be  in  equilibrium  around  this 
point. 

Now  the  force  Q,  whose  direction  is  tangent  to  the 
arc  FG,  and  which  we  may  regard  as  applied  to  the  point 
F  on  its  direction,  tends  to  turn  this  arc  in  the  same 
manner  as  the  force  M  (31),  and  may  be  substituted  for 
this  latter  force  to  counter-balance  the  force  P :  hence 
the  two  forces  P,  Q,  will  also  be  in  equilibrium  around 
the  fixed  point  D  ;  hence  their  resultant  will  be  de 
stroyed  by  the  resistance  of  this  point,  and  conse 
quently,  (31)  the  direction  of  this  resultant  will  pass 
through  the  point  D. 

But  the  resultant  of  the  two  forces  P,  Q,  should  pass 
through  the  point  of  intersection  A  of  their  directions 
(4) ;  hence  this  resultant  will  be  in  the  direction  of  the 
diagonal  AD. 


COMPOSITION   OF    FORCES. 


37 


33.  ANOTHER  DEMONSTRATION.* 
Suppose  that  the  force  Q  acts  at  the 
point  c  on  its  direction,  being  inva 
riably  fixed  at  the  point  A,  and 
that  there  is  applied  to  the  same 
point  c,  and  in  the  opposite  direc 
tions  CM,  CM',  two  forces  M,  M', 
each  equal  to  Q,  the  effect  of  the 
two  forces  P  and  Q  will  be  the  same 
as  that  of  the  four  forces  P,  Q,  M,  M',  since  the  last  two, 
M  and  M',  destroy  each  other,  being  equal  and  opposed. 
Now,  these  four  forces  form  two  couples ;  the  one,  Q  and 
M  intersect  at  the  point  c ;  the  other  is  composed  of  two 
parallel  and  unequal  forces  P,  M',  applied  to  the  line 
AC;  the  resultant  of  the  two  equal  forces  Q,  M,  is  di 
vided  along  the  line  CK,  which  divides  (12)  the  angle 
MCQ  into  two  equal  parts ;  the  resultant  HK  of  the  two 
forces  P,  M',  applied  to  the  points  A,  c,  of  the  line  AC, 
passes  through  a  point  H  of  this  line,  so  that  we 
have  (18) 

p  :  M',  or  Q  : :  HC  :  HA  ; 

moreover,  it  is  parallel  to  the  direction  AB  of  the  force 
p.  Hence  the  point  K,  the  intersection  of  the  two  re 
sultants  CK,  HK,  is  a  point  of  the  resultant  of  the  four 
forces  P,  Q,  M,  M',  and  consequently  of  the  first  two  P 
and  Q.  The  point  K  is  upon  the  direction  of  the  di 
agonal  AD  of  the  parallelogram  constructed  upon  AB  and 
AC,  as  sides ;  thus,  by  construction,  the  angle  HCK  is 
equal  to  the  angle  KCD.  Now,  the  angles  KCD  and  CKH 

*  This  demonstration  differs  but  little  from   that  given  by  M. 
Poirisot,  m  the  first  edition  of  his  Statique,  in  the  year  1803. 

4 


38  STATICS, 

are  equal,  being  alternate  internal  angles  ;  hence,  in  the 
triangle  CHK,  the  angles  K  and  o  are  equal ;  hence  it 
follows,  that  the  line  KII  has  the  same  length  as  the  line 
HC  ;  but  we  have  the  proportion  : 

p  :  Q  :  :  no  :  HA  ; 
hence  we  will  have 

p  :  Q  : :  KH  :  HA, 
and  since 

p  :  Q  :  :  CD  :  AC  , 

the  ratio  of  the  lines  AC  and  CD  is  the  same  as  that  of 
the  lines  AH  and  UK  ;  hence  the  three  points  A,  K,  D,  are 
in  a  straight  line,  and  this  line  is  the  diagonal  of  the 
parallelogram  constructed  upon  AB  and  AC  as  sides. 

COROLLARY  I. 

34.  If  from  any  point  D  (Fig.    9),  taken  upon  the 
direction  AD  of  the  resultant  of  the  two  forces  P,  Q,  the 
lines  DB,  DC  be  drawn  parallel  to  the  directions  of  these 
forces,  a  parallelogram  ABCD  will  be  formed,  whose  sides 
AB,  AC,  will  be  proportional  to  the  forces  P,  Q,  that  is  to 
say,  we  will  have  : 

p  :  Q  : :  AB  :  AC,  or  : :  DC  :  DB. 

For  if  these  sides  were  not  proportional  to  the  forces, 
their  resultant  would  be  in  the  direction  of  the  diagonal 
of  the  parallelogram  whose  sides  would  be  proportional 
to  these  forces  (32),  and  not  in  the  direction  AD,  which 
would  be  contrary  to  the  supposition. 

COROLLARY  II. 

35.  If  from  any  point  D  (Fig   9),    taken  upon  the 
direction  AD  of  the  resultant  of  the  two  forces  P,  Q, 


COMPOSITION   OF   FORCES.  39 

the  perpendiculars,  DE,  DF,  be  drawn  upon  the  directions 
of  these  two  forces,  these  perpendiculars  will  be  to  each 
other  reciprocally  as  the  forces  P,  Q. 
For  we  have  just  seen  (34)  that 

p  :  Q  :  :  DC  :  DB  ; 

and  the  similar  triangles  DEE,  DCF  give 
DC  :  DB  : :  DF  :  DE  ; 

hence 

p  :  Q  : :  DF  :  DE. 

THEOREM. 

36.    When  the  directions  of  the  two  forces  P,  Q,  are 
contained  in  the  same  plane,  and  coincide  in  a  point 
A,   if  the  lines  AB,  AC,  be  laid  off  on  these  directions 
proportional  to  these  forces,  so  that 
p  :  Q  :  :  AB  :  AC, 

and  the  parallelogram  ABDC  be  completed,  the  resultant 
R  of  these  two  forces  will  be  represented  in  intensity 
and  direction  l>y  the  diagonal  AD  of  the  parallelogram  ; 
that  is  to  say,  we  shall  have  : 

p  :  Q  :  R  : :  AB  :  AC  :  AD. 

Fig.  11. 

DEMONSTRATION.     We  have 

/  \  already  seen  (32)  that  the  re 

sultant  of  the  two  forces  P,  Q, 
will  be  in  the  direction  of  the. 
diagonal  AD  of  the  parallelo 
gram  ;  it  is  only  necessary  to 
show  that  its  intensity  will  be 
represented  by  this  diagonal. 


\ 


40  STATICS. 

Let  there  be  applied  to  the  point  A,  a  force  s,  equal 
and  directly  opposed  to  the  resultant  R ;  this  force  will 
be  in  the  direction  of  the  prolongation  of  the  diagonal 
DA,  and  the  three  forces  P,  Q,  S,  will  be  in  equilibrium. 
Hence  the  force  Q  will  also  be  equal  and  directly  opposed 
to  the  resultant  of  the  two  other  forces  P,  S ;  and  con 
sequently  this  last  resultant  will  be  in  the  direction  of 
the  prolongation  of  the  line  CA.  Let  CA  be  laid  off  on 
its  prolongation  from  A  to  H ;  draw  the  line  HB,  which 
will  be  parallel  to  AD,  and  consequently  to  the  direction 
of  the  force  s ;  and  through  the  point  H  draw  HK  parallel 
to  the  direction  of  the  force  P :  the  two  forces  P,  s,  will 
be  to  each  other  as  the  sides  AB,  AK,  of  the  parallelo 
gram  ABHK  (34),  that  is  to  say, 

p  :  s  : :  AB  :  AK  or  IIB. 

Now,  by  reason  of  the  parallelogram,  AD=BII  ;  more 
over,  the  two  forces  S  and  n  are  equal ;  hence  we  have, 

p  :  R  : :  AB  :  AD. 
But  by  supposition 

p  :  Q  : :  AB  :  AC. 

Hence,  by  combining  these  last  proportions,  we  will  have 
p  :  Q  :  R  : :  AB  :  AC  :  AD. 


COROLLARY  I. 

37.  If  the  two  forces  P,  Q,  be  applied  to  the  point  A, 
equilibrium  will  be  produced  by  applying  to  the  same 
point  a  third  force  in  the  direction  AK,  and  proportional 
to  the  diagonal  AD  ;  for  this  force  will  be  equal  and  di 
rectly  opposed  to  the  resultant  of  the  two  forces  P,  Q. 


COMPOSITION   OF    FORCES. 


41 


If  the  forces  P,  Q  be  applied  to  other  points  of  their 
direction,  equilibrium  can  be  produced  by  applying  to 
any  point  of  the  line  AD,  and  in  the  direction  DA,  a  force 
proportional  to  DA,  provided  the  point  of  application  of 
this  last  force  is  connected  in  an  invariable  manner 
with  the  points  of  application  of  the  two  forces  P,  Q. 


Fig.  11. 


COROLLARY  II. 


38.  A  force  R,  of  given  in 
tensity  and  direction,  may  be 
always  decomposed  into  two 
other  forces  P,  Q,  in  the  direc 
tion  of  the  given  lines  AP,  AQ, 
provided  these  directions  and 
that  of  the  force  R  are  included 
in  the  same  plane  and  coincide 
in  the  same  point  A. 


For  this  purpose,  we  will  represent  the  force  R  by  a  part 
AD  of  its  direction ;  then  drawing  through  the  point  D 
the  lines  DC,  DB,  parallel  to  the  given  directions  AP,  AQ, 
we  will  form  a  parallelogram  ABCD,  whose  sides  AB,  AC, 
will  represent  the  required  forces  P,  Q ;  for  (36)  the  re 
sultant  of  these  two  forces  will  have  the  same  intensity 
and  the  same  direction  as  the  force  R. 

The  intensities  of  the  two  forces  P,  Q,  may  be  found 
by  means  of  the  proportions : 

AD  :  AB  : :  R  :  P. 


AD  :  AC  : :  R  :  Q. 


42 


STATICS. 


Fig.  11. 


COROLLARY  III. 

39.  In  the  triangle  ABD 
(Fig.  11),  the  sides  AB,  BD,  AD, 
are  proportional  to  the  sines 
of  the  angles  CAD,  DAB,  BAG, 
formed  by  the  directions  of  the 
forces  P,  Q,  R  ;  hence  it  follows, 
that  we  shall  have  the  propor 
tion 


A          .        A  A 

p  :  Q  :  R  :  :  sin  (Q.R)  :  sin  (P.R)  :  sin  (P.Q) 

the  notation  (Q.R),  &c.,  being  the  angle  of  the  two  forces 
Q  and  R,  &c. 

The  same  triangle  ABD  gives 


hence 


COS  ABD; 


j^2pQ  cos  (P.Q). 


PROBLEM. 

40.  To  determine  the  resultant  of  any  number  of 
forces  P,  Q,  R,  S  .  .  .  .  whose  directions,  contained  or 
not  contained  in  the  same  plane,  meet  in  the  same 
point  A. 

SOLUTION.  Lay  off  upon  the  directions  of  all  the 
forces,  from  the  point  A,  the  lines  AB,  AC,  AD,  AE  .  .  .  pro 
portional  to  their  intensities ;  then,  considering  first  any 
two  of  these  forces,  as  P,  Q,  complete  the  parallelogram 
ABFC,  whose  diagonal  AF  will  represent  in  intensity  and 
direction  the  partial  resultant  T  of  these  two  forces  (36). 


COMPOSITION    OF   FORCES. 


43 


Fl'ff- 12-  Instead  of  the  forces 

p,  Q,  take  their  resultant 
T,  and  considering  the 
two  forces  T,  R,  complete 
the  parallelogram  AFGD, 
whose  diagonal  AG  will 
represent  in  intensity  and 
direction  the  resultant 
V  of  the  two  forces  T,  R, 
which  will  be  that  of  the 
three  forces  P,  Q,  R. 

In  like  manner,  instead 
of  the  forces  P,  Q,  R,  take 
their  resultant  v,  and 
considering  the  two  forces  V,  S,  complete  the  parallelo 
gram  AGUE,  whose  diagonal  AH  will  represent,  in  intensity 
and  direction,  the  resultant  x  of  the  forces  v,  s,  which 
will  also  be  that  of  the  four  forces  P,  Q,  R,  s. 

By  continuing  in  this  way,  the  direction  and  intensity 
of  the  general  resultant  of  all  the  forces  P,  Q,  R,  s  .  .  . 
will  be  found,  whatever  may  be  their  number. 

COROLLARY. 

41.  If  all  the  forces  P,  Q,  R,  s  .  .  .  .  be  applied  to  the 
point  of  coincidence  A  of  their  directions,  in  order  to 
produce  an  equilibrium,  find  first  the  intensity  and  di 
rection  of  their  resultant  (40) ;  then  apply  to  the  point 
A  a  force  equal  and  directly  opposed  to  it.  But  if  the 
forces  be  applied  to  other  points  of  their  directions, 
connected  together  in  an  invariable  manner,  an  equi 
librium  will  be  produced  by  applying  to  any  invariable 
point  in  the  direction  of  their  resultant,  a  force  equal 


44 


STATICS. 


and  directly  opposed  to  this  resultant,  provided  the 
point  of  application  of  this  force  is  also  connected  in  an 
invariable  manner  with  the  points  of  application  of  the 
forces  p,  Q,  E,  S  .  .  .  . 

PROBLEM. 

42.  To  determine  the  resultant  of  any  number  of 
forces,  whose  directions  included  in  the  same  plane  do 
not  meet  in  the  same  point,  ivhose  points  of  application 
A,  B,  c,  D  .  .  .  .  are  connected  together  in  an  invariable 
manner,  and  whose  intensities  are  represented  by  the 


parts  Aa, 


cc,  i)d, 

Fig.  13. 


of  their  directions. 


SOLUTION.  Hav 
ing  prolonged  the  di 
rections  of  any  two 
of  the  forces,  for  in 
stance  P,  Q,  until  they 
meet  somewhere  in  a 
point  E,  lay  off  from 
E  to  F  and  from  E  to 
G  the  lines  Aa,  B#, 
which  will  represent 
these  forces ;  and 
complete  the  paral 
lelogram  EFeG,  whose 
diagonal  ~&e  will  re 
present  in  intensity 
and  direction  the  re 
sultant  T  of  the  two 
forces  P,  Q  (86). 


COMPOSITION   OF   FORCES.  45 

Instead  of  the  forces  p,  Q,  take  their  resultant  T,  and 
prolong  its  direction  as  well  as  that  of  the  force  R  until 
they  meet  somewhere  in  a  point  H ;  lay  off  the  line  Ee 
from  H  to  I,  the  line  cc  from  n  to  K,  and  complete  the 
parallelogram  HI^K,  whose  diagonal  ilk  will  represent  in 
intensity  and  direction  the  resultant  v  of  the  two  forces 
T,  R,  which  will  be  also  that  of  the  three  forces  P,  Q,  R. 

In  like  manner,  instead  of  the  three  forces  P,  Q,  R, 
take  their  resultant  v,  and  prolong  its  direction,  as  well 
as  that  of  the  force  S,  until  they  meet  in  a  point  L ; 
then  laying  off  from  L  to  M  and  from  L  to  N  the  lines 
H^,  vd,  which  represent  the  forces  V  and  s,  complete 
the  parallelogram  LM?N,  whose  diagonal  Ll  will  represent 
the  resultant  x  of  these  two  forces,  which  will  also  be 
that  of  the  four  forces  P,  Q,  R,  s. 

By  thus  continuing,  the  intensity  and  direction  of  the 
general  resultant  of  all  the  proposed  forces  will  be 
found,  whatever  may  be  their  number. 

COROLLARY. 

43.  Hence,  when  several  forces,  directed  in  the  same 
plane,  are  applied  to  points  connected  together  in  an 
invariable  manner,  these  forces  always  have  one  result 
ant  :  thus  it  is  possible  to  produce  an  equilibrium  by 
means  of  a  single  force,  except  in  the  case  where  the 
direction  of  one  of  these  forces  being  parallel  to  that 
of  the  resultant  of  all  the  others,  the  force  and  the  re 
sultant  would  always  be  equal  to  each  other  and  would 
act  in  contrary  directions ;  for  we  have  seen  (26)  that, 
in  order  to  produce  an  equilibrium  in  that  case,  it  would 
be  necessary  to  apply  a  zero  force  in  a  line  situated  at 
an  infinite  distance ;  which  is  impracticable. 


46  STATICS. 


THEOREM. 

44.  If  three  forces  P,  Q,  R,  be  represented  in  intensi 
ty  and  direction  by  the  three  sides  AB,  AC,  AD,  adjacent 
to  the  same  angle  of  a  parallelopipedon  ABFEGD,  so 
that 

p  :  Q  :  R  : :  AB  :  AC  :  AD, 

their  resultant  s  will  be  represented  in  intensity  and  di 
rection  by  the  diagonal  AE  of  the  parallelopipedon  ad 
jacent  to  the  same  angle,  and  we  shall  have 
p  :  Q  :  R  :  s  :  :  AB  :  AC  :  AD  :  AE. 
DEMONSTRATION.     In  the  plane  ABCF,  which  contains 
the  directions  of  the  two  forces  P,  Q,  draw  the  diagonal 
Fig- u-  AF  ;  also  draw  the   diagonal 

DE  in  the  opposite  face  DHEG  : 
these  two  diagonals  will  be 
equal  and  parallel ;  for  the 
two  sides  AD,  EF  of  the  paral 
lelopipedon  at  the  extremi- 
ties  of  which  they  terminate 
are  parallel  and  equal :  hence 
AFED  will  be  a  parallelogram.  This  done,  the  two 
forces  P,  Q,  being  represented  in  intensity  and  direction 
by  the  sides  AB,  AC,  of  the  plane  ABFC,  which  is  a  paral 
lelogram,  their  resultant  T  will  be  represented  in  mag 
nitude  and  direction  by  the  diagonal  AF  (36),  and  we 
shall  have 

p  :  Q  :  T  : :  AB  :  AC  :  AF. 

Likewise  the  two  forces  T,  R,  being  represented  by  the 
sides  AF,  AD,  of  the  parallelogram  AFED,  their  resultant 


COMPOSITION    OF   FORCES.  47 

S,  which  will  be  also  that  of  the  three  forces  P,  Q,  R, 
will  be  represented  by  the  diagonal  AE  of  the  same  par 
allelogram,  and  we  shall  have 

T  :  R  :  S  : :  AF  :  AD  :  AE. 

Hence,  combining  the  two  proportions,  we  will  obtain 
p  :  Q  :  R  :  S  : :  AB  :  AC  :  AD  :  AE. 

Now  the  diagonal  AE  is  also  the  diagonal  of  the  par- 
allelopipedon ;  hence  the  resultant  of  the  three  forces 
p,  Q,  R  will  be  represented  in  intensity  and  direction  by 
the  diagonal  of  the  parallelopipedon. 

COROLLARY  I. 

45.  A  force  S  of  given  intensity  and  direction  can 
always  be  decomposed  into  three  other  forces  P,  Q,  R, 
in  the  direction  of  the  given  lines  AP,  AQ,  AR,  not  in 
cluded  in  the  same  plane,  provided  these  three  directions 
and  that  of  the  force  S  coincide  in  the  same  point  A. 

Thus,  through  the  three  directions,  considered  two 
and  two,  draw  the  three  planes  BAG,  CAD,  DAB  ;  repre 
sent  the  force  s  by  a  part  AE  of  its  direction ;  and 
through  the  point  E  draw  three  other  planes  EGDH, 
EHBF,  EFCG,  respectively  parallel  to  the  first  three. 
These  six  planes  will  be  the  faces  of  a  parallelopipedon, 
of  which  AE  will  be  the  diagonal,  and  of  which  the  sides 
AB,  AC,  AD,  taken  upon  the  three  given  directions,  will 
represent  the  intensities  of  the  three  required  forces  P, 
Q,  R  ;  for  (44)  the  resultant  of  these  three  forces  will 
have  the  same  intensity  and  direction  as  the  force  S. 

Or  else,  draw  through  the  point  E  three  lines  paral 
lel  to  the  directions  AP,  AQ,  AR  ;  and  the  parts  EF,  EH, 
EG,  of  these  sides,  included  between  the  point  E  and  the 
planes  BAG,  CAD,  DAB,  will  represent  the  intensities  of 


48  STATICS. 

the  required  forces  P,  Q,  R  ;  for  these  lines,  being  three 

'  sides  of  the  parallelopipedon,  are  respectively  equal  to 

the  other  sides  AB,  AC,  AD,  which  are  parallel  to  them. 


COROLLARY  II. 

46.  When  the  three  forces  P,  Q,  R,  are  perpendicular 
to  each  other,  the  resultant  s  is  the  diagonal  of  a  rectan 
gular  parallelopipedon,  whose  three  sides  adjacent  to 
the  same  summit  of  an  angle  are  equal  to  the  three 
forces  P,  Q,  R  ;  the  intensity  of  this  resultant  is,  in  this 
case,  expressed  by 


COROLLARY  III. 

47.  Whatever  may  be  the  number  of  forces  P,  Q,  R, 
s,  .  .  .  applied  to  the  fixed  points  A,  B,  c,  D,  .  .  .  we  can 
always  conceive  the  system  of  three  right  lines,  perpen 
dicular  to  each  other,  to  be  transferred  to  the  points  of 
application  of  the  forces  so  as  to  occupy  positions  paral 
lel  to  their  former  positions,  and  each  of  these  forces  to 
be  decomposed  into  three  others,  in  the  directions  of  the 
three  rectangular  lines  passing  through  the  point  of  ap 
plication  ;  then  all  the  forces  P,  Q,  R,  S,  .  .  .  will  be 
decomposed  into  three  systems  of  forces,  so  that  all  the 
forces  of  the  same  system  will  be  reduced  to  a  single 
force  along  the  same  direction ;  hence  all  the  forces  P, 
Q,  R,  s,  will  have  three  resultants  parallel  to  the  three 
rectangular  lines,  fixed  and  determined  in  position  with 
reference  to  these  forces.  (See  No.  53). 


COMPOSITION   OF   FORCES.  49 


COEOLLARY    IV. 

48.  Call  s,  s',  s"  .  .  .  .  the  forces  which  act  upon  a 
determined  point ;  and  drawing  through  this  point  three 
lines  fixed  and  perpendicular  to  each  other,  each  of  the 
forces  s,  s',  s",  .  .  .  will  be  decomposed  into  three  others 
p,  q,  r,  in  the  direction  of  the  rectangular  lines. 

In  like  manner  calling  p\  q',  rf,  the  three  component 
forces  of  the  force  s' ;  and  p",  q"9  r",  the  three  com 
ponent  forces  of  the  force  s",  &c. ;  the  resultant  of  all 
the  forces  S,  s',  s"  will  be  the  diagonal  of  a  rectangular 
parallelopipedon,  whose  three  sides  adjacent  to  the  same 
angle  will  be, 

For  the  first,  p+p'+p"+ ; 

For  the  second,  q+ q'+q"-\-.  .  .  .  ; 

For  the  third,  r+r'+r"+ 

Hence  the  expression  for  this  resultant  will  be 


THEOREM. 

49.  Two  forces  in  the  direction  of  lines  which  do 
not  meet  cannot  be  reduced  to  a  single  force  equivalent 
to  them. 

DEMONSTRATION.  Let  P  and  Q  be  the  two  forces 
whose  directions  do  not  meet.  If  a  third  force  R  causes 
an  equilibrium  to  subsist  between  them,  any  two  fixed 
points,  one  being  taken  upon  the  direction  of  this  force 
R,  and  the  other  upon  the  direction  of  the  force  P,  will 
necessarily  destroy  the  force  Q :  now  these  two  points 


50  STATICS. 

cannot  be  so  taken  that  the  line  which  unites  them  will 
not  meet  the  force  Q ;  hence  this  force  will  not  be  de 
stroyed  ;  hence  it  is  absurd  to  suppose  that  the  two  forces 
p  and  Q  can  have  a  single  resultant  R. 


THEOREM. 

50.  All  the  forces  P,  Q,  R,  s,  .  .  .  applied  to  the 
points  A,  B,  c,  D,  .  .  .  joined  together  in  an  invariable 
manner.,  may  in  general  be  reduced  to  tivo  forces  in  the 
directions  of  lines  which  do  not  meet. 

DEMONSTRATION.  Having  extended  the  lines  in  the 
directions  of  the  forces  P,  Q,  R,  .  .  .  until  they  cut  a 
plane  having  a  fixed  and  determined  position  with 
reference  to  these  lines,  we  may  consider  the  points  of 
intersection  as  the  points  of  application  of  the  forces ; 
now  each  force  may  be  decomposed  into  two,  one  situ 
ated  in  the  plane  and  the  other  perpendicular  to  this 
plane :  all  the  forces  directed  in  the  plane  will  have  one 
resultant;  the  forces  perpendicular  to  the  plane,  and 
consequently  parallel  to  each  other,  will  have  another 
resultant.  In  some  particular  cases,  these  two  resultants 
meet,  and  all  the  proposed  forces  P,  Q,  R,  S,  .  .  .  will  be 
reduced  to  a  single  one ;  but  in  general  they  will  not 
meet :  hence  we  shall  have  two  forces,  one  situated  in  a 
plane  assumed  arbitrarily,  and  the  other  perpendicular 
to  this  plane,  which  will  produce  an  equilibrium  with 
any  number  of  forces  P,  Q,  R,  s,  .  .  .  .  applied  to  the 
points  A,  B,  c,  D  .  .  .  .  It  is  necessary  to  except  from 
this  general  conclusion  the  particular  case  which  we  are 


COMPOSITION    OF    FORCES.  51 

about  to  examine,  and  which  takes  place  when  the  forces 
situated  in  the  plane  and  the  forces  perpendicular  to 
the  plane,  are  reduced  to  one  or  more  couples  of  equal 
and  parallel  forces  applied  in  opposite  directions  to  the 
same  right  line. 

COROLLARY. 

51.  When  two  forces  act  along  the  direction  of  lines 
which  do  not  meet,  there  are  an  infinite  number  of  sys 
tems  of  two  forces  acting  in  the  direction  of  other  lines 
which  do  not  meet,  whose  action  is  equivalent  to  that  of 
the  first  two  forces :  in  fact,  any  force  may  be  decom 
posed  into  two  other  forces,  one  perpendicular  to  a 
plane  assumed  at  will  and  the  other  situated  in  this 
plane ;  hence  any  two  forces  are  equivalent  to  two  other 
forces,  one  being  situated  in  the  plane  assumed  arbi 
trarily,  and  the  other  perpendicular  to  this  plane. 

PROBLEM. 

52.  Two  forces  being  given  in  the  direction  of  lines 
which  do  not  meet,  to  find  tivo  others  equivalent  to  them, 
one   of  which   is  in    the    direction  of  a   line  having 
a  given  position. 

SOLUTION.  Let  P  and  Q  be  the  two  given  forces ; 
having  drawn  a  plane  perpendicular  to  the  line  of  given 
position,  decompose  the  forces  P  and  Q  into  two  others 
P'  and  Q',  one  having  its  direction  in  this  plane  and  the 
other  perpendicular  to  the  same  plane  ;  decompose  the 
force  Q',  parallel  to  the  given  line,  into  two  others  q',  q'f, 
parallel  to  Q',  of  which  the  one  q'  will  pass  through  the 


52  STATICS. 

given  line,  and  the  other  q"  through  a  point  of  the  force 
P' ;  the  two  forces  P',  q",  meeting  in  the  same  point, 
will  be  reduced  to  a  single  force  q ;  hence  the  two  forces 
P  and  Q  will  be  transformed  into  two  other  equivalent 
forces  q,  q',  the  latter  of  which,  </,  will  pass  through  a 
given  line. 

53.  Examination  of  a  particular  case  of  the  compo 
sition  of  forces  applied  to  given  points  A,  B,  c,  D,  .  .  . 
invariably  joined  together. 


PROPOSITIONS. 

Having  decomposed  each  of  the  forces  P,  Q,  R,  s,  .  .  . 
applied  to  the  points  A,  B,  c,  D  .  .  .  into  two  others,  one 
situated  in  a  given  plane  K,  and  the  other  perpendicular 
to  this  plane,  let  S  and  T  be  the  resultants  of  these  two 
systems  of  forces ;  it  may  happen  that  the  first  system  of 
forces,  instead  of  having  a  single  force  S  for  resultant, 
will  only  reduce  to  a  couple  of  forces  +s,— s,  equal,  paral 
lel,  opposite,  and  applied  to  the  same  line ;  in  this  case, 
the  three  forces,  T,  +  s,— s,  will  be  equivalent  to  two  forces 
beyond  the  plane  of  the  couple  +s,  —  s:  likewise,  if 
instead  of  a  single  resultant  T,  the  forces  perpendicular 
to  the  plane  K  should  reduce  to  a  single  couple  +£,  —  t, 
(designating  by  this  expression  two  equal  and  parallel 
forces,  opposed  and  applied  to  the  same  line),  the  three 
forces  s,-f  t,— t,  will  also  be  equivalent  to  two  forces,  as 
in  the  preceding  case  :  finally,  if  all  the  forces  acting  in 
the  plane  K  and  perpendicularly  to  this  plane,  should  be 
reduced  to  two  couples  +s,— s,  and  -K,— £,  these  two 
couples  wrould  compose  a  single  one. 


COMPOSITION    OF   FORCES.  58 

The  three  propositions  which  have  just  been  an 
nounced  are  included  in  the  two  following : 

1st.  A  force  T  and  a  couple  +«,  — 8  are  equivalent  to 
two  forces  in  the  directions  of  lines  which  do  not  meet ; 

2d.  Two  couples  -fs,  —  s  and  +£,  —t  are  equivalent 
to  a  single  couple  of  a  like  nature :  such  for  example  as 
-f  r,  —  r.  ' 

Demonstration  of  these  two  Propositions. 

54.  IST  PROPOSITION.    A  force  T  and  a  couple +sy—s 
will  combine  into  two  forces :  thus,  the  plane  of  the 
couple  being  prolonged  intersects  the  force  T  in  a  point 
which  may  be  regarded  as  the  point  of  application  of 
the  force  T ;  drawing  any  line  through  this  point  and  in 
the  plane  of  the  couple,  and  regarding  this  line  as  fixed 
with  reference  to  the  three  forces  T,+s,— s  which  are 
applied  to  it,  decompose  the  force  T  into  two   other 
parallel  ones  t,  t'9  which  will  have  upon  the  fixed  line 
the  same  points  of  application  as  the  forces  -fs,— s. 
The  forces  £,  s,  meeting  in  the  same  point,  will  have  a 
resultant ;  so  also  will  the  forces  £',  —s :  they  will  have 
a  second  resultant ;  these  two  resultants  evidently  will 
be  equivalent  to  the  three  forces  T,+S, — s. 

55.  If  the  plane  of  the  couple  were  parallel  to  the 
force  T,  it  would  be  necessary  to  decompose  the  forces 

Fig.  13.  («.)  Of  the  couple+s  and  —  s  parallel  to  T. 
Suppose  this  couple  is  applied  to  the 
line  AB  (Fig.  13  a)  perpendicular  to 
this  line  ;  drawing  through  their  points 
of  application  A  and  B,  the  lines  AM, 
BN,  parallel  to  the  direction  of  the  force 
T,  decompose  the  force  +s  into  two 


54  STATICS. 

others  in  the  directions  AM  and  AB,  the  force  —  s  into 
two  others  BN  and  BA  ;  the  forces  along  the  direction 
AB  will  be  destroyed;  the  force  along  the  direction  AM 
or  BN,  and  the  force  T  which  is  parallel  to  it,  will  com 
bine  to  form  a  single  force  :  hence,  in  this  case,  the  three 
forces  T, +«,—«,  will  reduce  to  two,  one  in  the  plane 
of  the  couple  +s, — s,  and  the  other  parallel  to  this 
plane. 

56.  2D  PROPOSITION.  Two  couples  +«,— sand  -H,— £, 
situated  in  any  planes  whatever,  will  combine  into  a 
single  couple+r,— r :  thus,  the  planes  of  these  couples 
being  prolonged  will  meet  in  the  direction  of  a  line 
which  may  be  considered  as  invariably  joined  to  the 
points  of  application  of  the  forces  which  compose  the 
two  couples.  Let  KL  (Fig.  13  5)  be  this  line,  the  in 
tersection  of  the  two  planes  LKMN,  LZM'N',  one  of  which 
contains  the  couple  -fs, — s,  applied  to  the  line  AB,  the 
other  the  couple  +£,— t,  applied  to  the  line  CD  ;  the 
directions  of  the  forces  +», — #>~M> — t  intersect  this 
line  in  the  points  a,  b,  c,  d ;  dividing  the  lines  aby  cd 
Fig.  13,  (b.)  int0  two  equal  parts,  and  marking  the 
middle  points  /',  /,  we  may  transfer 
the  couple  -H,— t,  applied  to  the  line 
CD,  parallel  to  itself,  so  that  the  points 
/,  /'  will  coincide  ;  we  will  then  have  'a 
new  couple  -M',— t'.  It  is  necessary,  in 
the  first  place,  to  demonstrate  that  this 
second  couple,  composed  of  forces  equal 
and  parallel  to  those  of  the  first  and 
applied  to  the  line  C'D'  equal  to  CD,  will  be  in  equili 
brium  with  it,  and  that  in  general  we  cannot  change  the 
condition  of  equilibrium  of  two  couples,  by  transferring 


M 


COMPOSITION    OF   FORCES.  55 

one  of  the  couples  parallel  to  itself  in  its  plane.  Now, 
this  proposition  is  evident,  for  the  point  o  being  the 
middle  of  the  line  c'd,  the  two  forces  +£',-—£,  as  well  as 
the  two  forces  — t'9-\-t,  which  act  at  equal  distances 
from  this  point  o,  are  in  equilibrium ;  hence  the  second 
couple  may  be  substituted  for  the  first.  Now,  the  force 
+tf  is  decomposable  into  two  other  parallel  forces  pass 
ing  through  the  points  a  and  b;  the  force  ~—t'  also  may 
be  decomposed  into  two  others  passing  through  the 
points  b  and  a;  and  since  c'a=dfb,  the  component 
forces  of  these  two  forces  -H'  and  — £',  will  be  equal,  and 
will  differ  only  in  direction :  hence,  each  of  the  three 
forces  meeting  in  the  point  a  will  be  equal  to  one  of  the 
three  forces  meeting  in  the  point  b  ;  hence,  the  result 
ants  of  the  two  systems  of  forces  applied  to  the  points 
a  and  b  will  be  equal  and  opposed ;  hence  it  follows 
that  the  two  couples  +s,— s  and  -H,— -£  will  be  re 
duced  to  a  single  one  4-r, — r. 

If  the  forces  +t,—t  were  parallel  to  the  line  LK,  the 
couple  -fs,  —  s  (55)  might  be  changed  into  another 
+«',—  s',  whose  forces  would  be  directed  parallel  to  the 
line  LK,  and  the  four  parallel  forces  -f£,  +  s', — £, — s' 
would  be  reduced  to  two  equal  and  opposed  forces 
+  (*+*')  and  -(£+<).* 


*  See  the  theory  of  couples  in  the  Statique  of  M.  Poinsot,  6th 
edition,  1834. 


56 


CHAPTER  SECOND. 

OP   MOMENTS. 

57.  Two  kinds  of  moments  are  to  be  considered.    The 
moment  of  a  force  referred  to  a  point  is  the  product  of 
this  force  multiplied  by  the  perpendicular  let  fall  from 
the  point  upon  the  force ;  the  moment  of  a  force  referred 
to  a  plane,  is  the  product  of  this  force  multiplied  by  the 
distance  of  its  point  of  application  from  the  plane :  this 
second  kind  of  moment  does  not  change,  even  though 
the  forces  vary  in  direction ;  they  differ  in  this  condi 
tion  from  the  moments  of  the  first  kind,  which  are  inde 
pendent  of  the  points  of  application  of  the  forces  whose 
direction  is  constant. 

58.  When  the  moments  of  several  forces,  referred  to 
the  same  point,  are  considered,  this  point  is  named  the 
centre  of  moments, 

59.  Hence  it  follows,  if  the  intensity  of  a  force  be 
knowTn,  and  its  moment  referred  to  a  centre  or  a  plane, 
and  if  the  plane  be  parallel  to  the  force,  we  will  obtain 
the  distance  of  the  centre,  or  the  plane,  from  the  di 
rection  of  the  force,  by  taking  the  quotient  of  the 
moment  divided  by  the  force ;  if  the  moment  and  the 
distance  be  known,  we  shall  obtain  the  intensity  of  the 
force  by  taking  the  quotient  of  the  moment  divided  by 
the  distance. 


MOMENTS. 


57 


15- 


COROLLARY. 

60.  When  the  two  parallel  forces  P,  Q,  act  in  the  same 
direction,  their  moments,  referred  to  any  point  c  in  the 
direction  of  their  resultant,  are  equal. 

For  if  through  the  point  c  the 
line  AB  be  drawn  perpendicular  to 
the  direction  of  the  two  forces,  and 
terminate  in  A  and  B,  this  line  will 
be  divided,  at  the  point  c,  into  two 
parts  reciprocally  proportional  to 
,the  forces  P,  Q  (18):  that  is  to  say, 
we  shall  have 

p  :  Q  : :  BC  :  AC. 

Hence,  the  product  of  the  extremes  being  equal  to 
the  product  of  the  means,  we  shall  have 
PXAC=QXBC. 


THEOREM. 

61.  If  at  the  extremities  A,  B  of  an  inflexible  right 
line,  two  parallel  forces  P,  Q  be  applied  which  act  in  the 
same  direction,  and  if  through  the  point  of  application 
c  of  their  resultant,  a  right  line  DB  be  drawn  in  any 
plane,  upon  which  the  perpendiculars  AD,  BE  be  dropped 
from  A  and  B,  ive  shall  have 

PXAD  =  QXBE. 


58 


STATICS. 


Fig.  15.  DEMONSTRATION.     The  right  an 

gled  triangles  ADC,  EEC,  which  are 
similar,  because  the  angles  opposite 
the  summit  c  are  equal,  give 

EC  :  AC  : :  BE  :  AD. 
ow  we  have  (18) 

p  :  Q  :  :  EC  :  AC. 
Hence 

p  :  Q  : :  BE  :  AD  ; 
and,  the  product  of  the  extremes  equalling  that  of  the 


means, 


PXAD  =  QXBE. 


THEOREM. 


Fig.  16. 


* 

f— AB 


62.  Two  parallel  forces  P,  Q, 
which  act  in  the  same  direc 
tion^  being  applied  to  the  points 
A,  B  of  an  inflexible  right  line, 
and  through  a  point  F  of  this 
line,  the  line  FH  being  drawn 
in  any  plane  : 

1st.  If  the  point  F  be  taken 
upon  the  prolongation  of  AB, 
and  if  from  the  points  A,  B, 
and  from  the  point  of  appli 

cation  c  of  the  resultant,  the  perpendiculars  AG,  BH,  ci, 

be  dropped  upon  FIT,  we  shall  have 

11XCI=QXBH  +  PXAG  ; 


MOMENTS. 


59 


Fi9-  17.  2d,  If  the  point  F  be  taken  be 

tween  A  and  B,  we  shall  have 
RXCI=QXBH  — PXAG. 

DEMONSTRATION.     Through  the 
point  c,  draw  DE  parallel  to  FH, 
intersecting  the  perpendiculars  AD, 
BH  in  the  points  D,  E  ;  we  have 
DG=CI=EII,  besides  (61),  PXAD=QXBE. 


Fig.  16. 

X 


Now,  in  the  first  case,  the 
resultant  R  being  equal  to  the 
sum  of  the  two  forces  P,  Q 
(18),  we  shall  have 

R  x  c  i  =  (Q  +  P)  x  ci, 
or 


But  we  have  GD=AD+AG; 
hence 

RXCI=QXHE+PXAD+PXAG; 
or  placing  the  product  QXBE, 
in  place  of  PXAD  which  is  equal  to  it,  we  shall  have  : 

RXci=QXiiE-f  QXBE+PXAG; 
or  finally 

RXCI=QXBH  +  PXAG. 

In  the  second  case  we  have  likewise 

RXCI=QXHE-fPXGD. 

But  we  have  GD=AD—  AG;  hence 

—  PXAG 


60 


STATICS. 


or  substituting  for  PXAD  its  value  QXBE,  we  shall  have 

RXCI=QXHE  +  QXBE— PXAG  ; 

and  finally, 

RXCI=QXBH— PXAG. 


COROLLARY  I. 


Fig.  16. 


\ 


V   '^  ^ 

JB-  *——*-•"--- 


\ 


63.  Hence  it  follows,  1st. 
When  the  two  points  A,  B  are 
on  the  same  side  of  the  line 
FH,  the  distance  ci  of  the  point 
c  of  the  resultant  from  this 
line  will  be,  since  R=p-f-Q, 

QXBII-fPXAG 


Cl 


P  +  Q 


. 

X 


Fi9- 17-  2d.  When  the  points  A,  B  are 

placed  on  different  sides  of  the 
line  FH,  this  distance  will  be 

_QXBH  —  PXAG 

P  +  Q 

In  this  case  the  point  c  will  be 
placed,  with  reference  to  the  line 
GH,  on  the  same  side  as  the  force  Q,  for  which  the  pro 
duct  QXBH  is  the  greater. 


MOMENTS. 


61 


Fig.  16. 


\  V 


and  for  Fig.  37, 


COROLLARY  II. 

64.  If  the  line  FH  is  per 
pendicular  to  AB,  the  lines 
AG,  BIT,  ci  will  all  three  be  in 
the  direction  of  AB,  and  the 
proposition  enunciated  in  the 
preceding  theorem  will  still 
take  place.  In  this  case  we 
shall  have 

AG=AF,  BH=BF,  CI=CF. 

Then   we   shall   have   for 
Fig.  16, 

RXCF=QXBF  +  PXAF ; 
RXCF=QXBF— PXAF. 

As  to  the  distance  CF  from  the 
point  F  to  the  point  of  applica 
tion  c  of  the  resultant  R=P+Q, 
it  will  be  for  Fig.  16, 

_QXBF  +  PXAF 

P+Q 
and  for  Fig.  17, 

_QXBF— PXAF 
P  +  Q 

THEOREM. 

65.  Tivo  parallel  forces  P,  Q  (Figs.  18  and  19), 
which  act  in  the  same  direction,  being  applied  to  the 
points  A,  B  of  an  inflexible  right  line,  and  having  drawn 
a  plane  MN,  through  a  point  F  of  this  line,  parallel  to 
their  directions : 


<- 

D 

1 

,''  G 
/' 

62 


STATICS. 


1st.  If  the  point  F  be 
taken  upon  the  prolongation 
of  AB,  the  sum  of  the  mo 
ments  of  the  two  forces  P,  Q, 
referred  to  the  plane  MN, 
will  be  equal  to  the  moment 
of  their  resultant  R :  that 
is  to  say,  if  from  the  points 
A,  B,  and  from  the  point  of 
application  c  of  the  result 
ant,  the  perpendiculars  AG, 
BH,  ci,  be  let  fall  upon  the  plane,  we  shall  have 

RXCI=QXBH  +  PXAG. 
Fig.  19. 

2d.  If  the  point  F  be  taken  be- 
tiveen  A  and  B,  the  difference  of 
the  moments  of  the  forces  P,  Q  will 
be  equal  to  the  moment  of  their 
resultant :  that  is  to  say,  we  shall 
have 

RXCT=QXEH  —  PXAG. 


DEMONSTRATION.  The  three  lines  AG,  BH,  ci,  per 
pendicular  to  the  same  plane  MN,  are  parallel  to  each 
other ;  moreover  they  pass  through  the  three  points 
A,  B,  c,  of  the  same  line :  hence  they  are  in  the  same 
plane  drawn  through  A,  B  ;  hence  their  feet  G,  H,  I,  and 
the  point  F  are  in  the  same  plane.  But  the  four  points 
F,  G,  H,  I,  are  also  in  the  plane  MN  ;  hence  they  are  in 
the  intersection  of  two  different  planes,  and  conse 
quently  in  a  straight  line.  Now  draw  the  line  FGIH  : 


MOMENTS.  03 

it  will  intersect  the  lines  AG,  BH,  ci,  at  right  angles ;  for 
it  will  be  in  the  plane  MN  to  which  these  lines  are  per 
pendicular,  and  it  will  pass  through  their  feet.  Hence, 
by  considering  FGIH  as  the  line  FH  (Figs  16  and  17) : 

1st.  When  the  point  F  (Fig.  18)  is  upon  the  prolonga 
tion  of  AB}  we  shall  have  (62) 

RXCT  =  QXBH  +  PXAG. 

2d.  When  the  point  F  (Fig.  19)  is  between  A  and  B, 
we  shall  have 

RXCT  =  Q  +  BH  — PXAG. 

COROLLARY. 

66.  Hence,  1st.  When  the  two  forces  P,  Q,  (Fig.  18) 
are  on  the  same  side  of  the  plane  MN,  the  distance  ci 
of  the  plane  from  the  resultant  will  be  equal  to  the  sum 
of  the  moments  of  the  forces  referred  to  the  plane,  di 
vided  by  the  resultant  R,  or,  what  is  the  same  (18),  di 
vided  by  the  sum  P+Q  of  the  forces :  that  is  to  say,  we 
shall  have 

_QXBH  +  PXAG 

P  +  Q 

2d.  When  the  plane  passes  between  the  directions  of 
the  two  forces  (Fig.  19),  this  distance  will  be  equal  to 
the  difference  of  the  moments  divided  by  the  sum  of  the 
forces :  that  is  to  say,  we  shall  have 

_QXBH  — PXAG 

CI — . 

P  +  Q 

In  the  latter  case,  the  resultant  will  be  situated  on 
the  same  side  of  the  plane  MN  as  the  force  whose  mo 
ment  is  the  greater. 


STATICS. 


THEOREM. 

67.  If,  to  any  number  of  points  A,  B,  c,  D,  .  .  .  situ 
ated  or  not  in  the  same  plane,  but  connected  together  in 
an  invariable  manner,  the  parallel  forces  P,  Q,  R,  s,  .  .  . 
be  applied,  which  act  in  the  same  direction,  and  which 
are  all  placed  on  the  same  side  of  any  plane  MIST  parallel 
to  their  directions,  the  sum  of  the  moments  of  all  the 
forces  referred  to  the  plane  MN  will  be  equal  to  the  mo 
ment  of  their  resultant. 

DEMONSTRATION.     Draw  the  line  AB,  and  let  E  be  the 
point  on  this  line  through  which  the  resultant  T  of  the 
Fig.  20.  two  forces  P,  Q  pass 

es  ;  draw  the  line 
EC,  and  let  F  be  the 
point  on  this  line 
through  which  the 
resultant  v  of  the 
two  forces  T,  R, 
passes,  which  will  al 
so  be  the  resultant 
of  the  three  forces 
p,  Q,  R ;  draw  FD, 
and  let  G  be  the 

point  on  this  line  through  which  the  resultant  x  of  the 
two  forces  v,  s,  passes,  which  will  also  be  the  resultant 
of  the  four  forces  p,  Q,  R,  S  ;  and  so  on.  Finally,  from 
the  points  A,  B,  c,  D  .  .  .  .  and  the  points  E,  F,  G,  .  .  . 
let  fall  upon  the  plane  MN  the  perpendiculars  Aa,  B&,  cc, 
vd,  .  .  .  .  ,  EC,  F/,  ag,  .  .  .  . 


MOMENTS.  65 

Then,  the  moment  of  the  resultant  T  will  be  equal  to 
the  sum  of  the  moments  of  its  two  components  P,  Q  (65), 
and  we  shall  have 


In  like  manner,  the  moment  of  the  force  v  will  be 
equal  to  the  sum  of  the  moments  of  its  two  components 
T,  K,  and  we  shall  have 


Hence,  substituting  for  TXEe  its  value,  we  get 

VXF/=PXAa-f  QXB#-f  RXCtf. 

Likewise  the  moment  of  the  force  x  will  be  equal  to 
the  sum  of  the  moments  of  its  two  components  V,  s, 
which  will  give 

X  X  G<7  =  V  X  F/-f  S  X  Dd. 

Hence,  substituting  for  vXF/its  value,  we  shall  have 


And  so  on,  whatever  may  be  the  number  of  forces. 
Hence  the  moment  of  any  resultant  is  equal  to  the  sum 
of  the  moments  of  all  the  components  ;  hence,  &c. 


COROLLARY  I. 

68.  We  have  seen  (27)  that  the  intensity  of  the  re 
sultant  x  of  the  forces  P,  Q,  R,  s,  .  .  .  .  is  equal  to  the 
sum  p+Q-fR  +  s  ....  of  these  forces:  hence  the  dis 
tance  G(/  of  the  direction  of  this  resultant  from  the 
plane  MN  is  equal  to  the  sum  of  the  moments  of  all  the 


66  STATICS. 

forces  P,  Q,  R,  s,  .  .  .  divided  by  the  sum  of  all  these 
forces :  that  is  to  say,  we  shall  have 


P+Q+R+S 


COROLLARY  II. 

69.  Hence,  if  an  indefinite  plane  be  drawn  on  the  side 
on  which  the  forces  are  placed,  parallel  to  MN  and  at  a 
distance  Gg  from  it  :  that  is  to  say,  equal  to 


P  +  Q-I-R  +  S 

this  plane  will  contain  the  direction  of  the  resultant  of 
all  the  forces  P,  Q,  R,  S  .  .  .  .  ;  for  this  plane  will  con 
tain  all  the  points  which,  on  this  side,  are  distant  from 
the  plane  MN  by  the  quantity  Gg,  consequently  all  the 
points  of  the  direction  of  the  resultant. 


COROLLARY  III. 

70.  If  the  forces  P,  Q,  R,  s,  .  .  .  be  situated  on  each 
side  of  the  plane  MN,  the  moment  of  their  result 
ant,  referred  to  this  plane,  will  be  equal  to  the  excess 
of  the  sum  of  the  moments  of  the  forces  which  are  situ 
ated  on  one  side  of  the  plane,  over  the  sum  of  the  mo. 
ments  of  the  forces  which  are  situated  on  the  other  side. 

Thus,  let  v  be  the  partial  resultant  of  all  the  forces 
P,  Q  .  .  .  which  are  situated  on  one  side  of  the  plane, 
whatever  their  number  may  be,  and  E  the  point  of  ap- 


MOMENTS. 


67 


plication  of  this  force.     In  like  manner,  let  x  be  the 
Fiff-  21-  partial  resultant  of 

all  the  forces  R,  s, 
.  .  .  which  are  situ 
ated  on  the  other 
side,  and  F  the 
point  of  application 
of  this  force.  By 
letting  fall  upon 
the  plane  the  per 
pendiculars  Aa,  %b, 
Ee  .  .  .  cc,  fid,  F/,  .  . 
we  have  just  seen 
(67)  that  we  shall 
have 


and 


Now  let  Y  be  the  resultant  of  the  two  forces  v,  x,  and 
G  its  point  of  application  ;  this  force  will  be  the  general 
resultant  of  all  the  forces  P,  Q,  K,  s,  .  . 

This  done,  the  two  forces  v,  x,  being  situated  on  each 
side  of  the  plane  MIST,  the  moment  of  their  resultant  is 
equal  to  the  difference  of  their  moments  (65) ;  hence, 
letting  fall  the  perpendicular  Gg  upon  the  plane,  we 
shall  have 

YXG#=VXEe  — XXF/. 

Hence,  by  substituting  the  values  of  these  last  two  mo 
ments,  we  shall  have 


STATICS. 


68 


Hence,  &c. 


COROLLARY  IV. 

71.  Hence,  in  general,  in  whatever  manner  several 
parallel  forces  P,  Q,  R,  s,  .  .  .  acting  in  the  same  direc 
tion,  may  be  situated  with  reference  to  a  plane  MN, 
parallel  to  their  directions,  the  distance  &g  of  their  re 
sultant  from  this  plane  is  equal  to  the  excess  of  the  sum 
of  the  moments  of  the  forces  situated  on  one  side  of  the 
plane,  over  the  sum  of  the  moments  of  the  forces  situ 
ated  on  the  other  side,  divided  by  the  sum  of  all  the 
forces :  that  is  to  say,  we  shall  have 

_pXAa-f-QXB&  .  .  . .— (nxcc+sxvd . .  .)> 
g~~  P-fQ+R  +  S 

and  this  resultant  is  placed,  with  reference  to  the  plane 
MN,  on  the  side  on  which  the  sum  of  the  moments  is  the 
greater. 


COROLLARY  V. 

72.  Hence,  if  on  the  side  of  the  plane  MN,  on  which 
the  sum  of  the  moments  is  the  greater,  we  draw  an  in 
definite  plane  parallel  to  it,  which  is  distant  by  the 
quantity  G^,  or 


P-f  Q-f  R  +  S  ... 


MOMENTS. 


this  plane  will  contain  the  direction  of  the  resultant  of 
all  the  forces  P,  Q,  R,  S,  .  .  . 


COROLLARY  VI. 


Fig.  22. 


.  c 


73.  If  the  directions 
of  the  forces  P,  Q,  R,  S, . . . 
be  all  situated  in  the  same 
plane,  perpendicular  to 
the  plane  MN,  the  lines 
Atf,  B&,  cc,  vd  .  .  .  .  Gg 
will  fall  upon  the  line  KL, 
the  intersection  of  the 
two  planes  ;  and  we  shall 
then  also  have 


according  as  the  forces  are  on  the  same  side  or  on  the 
opposite  sides  of  the  line  KL.  Hence  we  shall  have,  in 
the  first  case,  Fig.  22, 


and,  in  the  second  case,  Fig.  23, 


P+Q+R+S 


70 


STATICS. 


23-  that  is  to  say,  when  several 

parallel  forces,  situated  in 
the  same  plane,  act  in  the 
same  direction,  the  distance 
of  their  resultant  from  any 
right  line,  traced  in  the 
same  plane  and  parallel  to 
their  directions,is  in  general 
equal  to  the  excess  of  the 
sum  of  the  moments  of  the 
forces  situated  on  one  side 
of  the  line,  over  the  sum 
of  the  moments  of  the  for 
ces  situated  on  the  other 
side,  divided  by  the  sum  of  the  forces. 


PROBLEM. 

74.  An  indefinite  number  of  parallel  forces  being 
given,  which  act  in  the  same  direction  and  whose  points 
of  application  may  or  may  not  be  situated  in  the  same 
plane,  to  determine,  by  means  of  moments,  the  direction 
of  the  resultant  of  all  these  forces. 

SOLUTION.  Having  drawn  at  pleasure  two  different 
planes  ABCD,  BCIK,  parallel  to  the  directions  of  the 
forces,  find  the  distance  of  the  resultant  from  each  of 
these  planes  separately  (71) ;  then  draw  a  plane  EFGH 
parallel  to  ABCD,  at  the  distance  from  this  latter  plane 
that  the  resultant  is  from  it,  and  situated  on  the  side 
on  which  the  sum  of  the  moments  referred  to  the  plane 
ABCD  is  the  greater ;  then  this  plane  EFGH  will  contain 


MOMENTS. 


71 


the  direction  required  (71).     Likewise  draw  a  plane 
•%•  24-  LMNO  parallel   to  BCIK, 

~ F 


at  a  distance  from  this 
latter  plane  equal  to  that 
of  the  resultant  from  it, 
and  situated  on  the  side 
on  which  the  sum  of  the 
moments  referred  to  the 
plane  BCIK,  is  the  great 
er ;  and  this  plane  will 
also  contain  the  direction 
required.  Hence  the  direction  of  the  resultant  being 
both  in  the  plane  EFGH,  and  in  the  plane  LMNO,  it  will 
be  in  the  line  of  intersection  PQ  of  these  two  planes. 


COROLLARY   I. 

75.  We  have  seen  (30)  that  if  several  parallel  forces 
change  in  direction,  without  changing  either  in  intensity 
or  their  points  of  application,  and  without  ceasing  to  be 
parallel  to  each  other,  their  resultant  always  passes 
through  the  same  definite  point,  which  has  been  termed 
the  centre  of  parallel  forces;  hence,  for  the  parallel 
forces  which  we  have  just  been  considering,  the  centre 
is  placed  in  the  direction  PQ  of  their  resultant. 

To  find  this  centre,  draw  at  pleasure  a  third  plane 
ABKR  (Fig.  24),  and  conceive  that  all  the  forces,  with 
out  changing  either  their  intensities  or  points  of  appli 
cation,  are  directed  parallel  to  each  other  and  to  the 
plane  ABKR  ;  find  the  distance  of  the  resultant  of  these 
new  forces  from  this  plane  (71).  This  done,  if  we  draw 
a  plane  STVX  parallel  to  ABKR,  and  at  a  distance  from 


72 


STATICS. 


this  latter  plane  equal  to  that  we  shall  have  found,  this 
plane  will  contain  the  new  resultant,  and  consequently 
the  centre  of  the  forces.  Hence  the  centre  of  the 
forces,  being  found  both  in  the  line  PQ  and  in  the  plane 
STVX,  it  will  be  found  in  the  point  of  intersection  Y  of 
the  line  and  the  plane ;  or,  which  is  the  same  thing, 
this  centre  will  be  found  at  the  point  of  intersection  Y 
of  the  three  planes  EFGH,  LMNO,  STVX. 

COROLLARY  II. 

76.  If  the  parallel  forces  P,  Q,  R,  S,  .  .  .  acting  in  the 
same  direction,  be  situated  in  the  same  plane ;  in  order 
Fig.  25.  (a)  to  find  the  position  of  their 

~  resultant,  draw  in  this  plane 
a  line  LN  parallel  to  the  direc 
tions  of  the  forces,  and  hav 
ing  let  fall  upon  this  line,  from 
all  the  points  of  application 
A,  B,  c,  D,  .  .  .  the  perpen 
diculars  A#,  B&,  c<?,  i>d  .  .  ., 
lay  off  upon  a  line  LM  per 
pendicular  to  LN,  the  line  L#', 
equal  to  (73) 


P+Q+R+S  .  .  . 

and  the  line  </'Y,  drawn  through  the  point  g'  parallel  to 
LN,  will  be  the  direction  of  the  resultant. 

If  all  the  forces  were  not  on  the  same  side  of  the  line 
LN,  it  would  be  necessary  to  subtract  the  moments  of 
the  forces  situated  on  the  other  side,  instead  of  adding 
them  (73). 


MOMENTS. 


73 


COROLLARY  III. 

77.  To  find  the  centre  of 
—  the  parallel  forces  P,  Q,  R,  s, 
..  ,,  .  included  in  the  same 
plane,  conceive  these  forces, 
without  changing  their  intensi 
ties  and  without  ceasing  to  be 
applied  to  the  same  points 
A,  B,  c,  D,  .  .  .  to  be  in  direc 
tions  parallel  to  another  line, 
as  LM,  upon  which  let  fall  the 
perpendiculars  Aa',  B£',  c<?', 

Ddf,  ....  and  the  distance  &gr  of  this  line  from  the  new 

resultant  will  be  (73) 

'+QXB£/-J-RXCC/ 


1         t 

Fig.  25.  ( 
t    <l    g      < 

9. 

:     i 

5            N 



A 

B 

_  _j  G-  „  j 

c 

A 

>4i 

Nfe*    /> 

R 

P-f  Q-fR  +  S  .  .  . 

Hence,  if  we  lay  off  upon  a  line  perpendicular  to  LM, 
the  line  *Lg  equal  to  this  distance,  and  through  the  point 
g  draw  g&  parallel  to  LM,  this  line  g&  will  be  the  direc 
tion  of  the  new  resultant.  Now,  the  centre  of  the  forces 
is  to  be  found  both  upon  the  direction  of  the  first  result 
ant  #'Y,  and  upon  that  of  the  second  g& ;  hence  it  will 
be  at  the  point  of  intersection  G  of  these  two  directions. 

If  all  the  forces  were  not  on  the  same  side  of  the  line 
LM,  it  would  be  necessary  to  subtract  the  moments  of 
those  which  are  situated  on  the  other  side,  instead  of 
adding  them. 


74  STATICS. 


COROLLARY  IV. 

78.  If  the  points  of  application  A,  B,  c,  D,  .  .  .  (Fig. 
25)  are  in  the  same  plane  to  which  the  directions  of  the 
parallel  forces  P,  Q,  R,  s,  .  .  .  are  oblique,  the  centre  G 
of  these  forces  will  also  be  in  this  plane  (30),  and  its 
position  will  be  the  same  as  though  the  directions  of  the 
forces  were  parallel  to  each  other  and  situated  in  this 
plane.  Thus,  to  find  in  this  case  the  centre  of  the 
forces  G,  draw  in  the  plane  any  two  right  lines  LN,  LM  ; 
then  suppose  that  the  forces  are  in  a  direction  parallel 
to  LN,  and  find  (77)  the  direction  g'Y  of  their  resultant 
on  this  supposition  ;  then  suppose  they  are  in  a  direction 
parallel  to  LM,  and  find  the  direction  Gg  of  their  re 
sultant;  the  point  of  intersection  G  of  the  two  lines 
<7'Y,  g&  will  be  the  centre  of  the  forces  required. 

The  centre  being  found,  if  we  draw  through  this 
point  a  line  parallel  to  the  real  directions  of  the  forces 
p,  Q,  R,  s,  .  .  .  this  line  will  be  the  direction  of  their 
resultant. 


COROLLARY  V. 

79.  Finally,  if  the  points  of  application  a',  bf,  c',  d', .  .  . 
(Fig.  25)  be  upon  the  same  straight  line  LM  oblique  to 
the  direction  of  the  forces,  the  centre  g'  of  these  forces 
will  be  upon  this  line  (30),  and  its  position  will  be  the 
same  as  though  the  directions  of  the  forces  were  per 
pendicular  to  LM. 


MOMENTS. 


75 


Hence   (76)  the  distance  g'~L  of  this  centre  from  a 
given  point  L  upon  the  line,  will  be  equal  to 


L  +  QX^'L-j-RXc 


P+Q+R  +  S  .  .  . 

If  all  the  forces  are  not  situated  on  the  same  side 
with  reference  to  the  point  L,  it  will  be  necessary  to 
subtract  the  moments  of  those  which  are  situated  on  the 
other  side,  instead  of  adding  them. 


Fig.  26. 


*r 


LEMMA. 

80.  When  the  directions  of 
two  forces  P,  Q,  meet  in  a  point 
A,  the  moments  of  these  forces,  re 
ferred  to  any  point  D  in  the  line 
of  direction  of  their  resultant  R, 
are  equal. 

For  we  have  seen  (35)  that  if 
from  the  point  D  the  perpendicu 
lars  DB,  DC,  be  dropped  upon  the 
directions  of  the  forces,  prolonged 
if  necessary,  we  shall  have 


p  :  Q  : :  DC  :  DB. 

Hence,  the  product  of  the  extremes  being  equal  to 
that  of  the  means,  we  shall  have 


PXDB=QXDC. 


76 


STATICS. 


COROLLARY. 

81.  From  this  it  follows  that,  if  the  directions  of  the 
two  forces  P,  Q  meet-  in  a  point  A,  the  moment  of  any 


Fig.  27. 


one,  Q,  of  these  forces,  referred  to 
a  point  D  in  the  direction  of  the 
other,  will  be  equal  to  the  moment 
of  their  resultant  R  referred  to  the 
same  point :  that  is  to  say,  by  let 
ting  fall  from  the  point  D,  the  per 
pendiculars  DB,  DC  upon  the  direc 
tion  of  the  force  Q,  and  upon  that 
of  the  resultant  R,  prolonged  if  ne 
cessary,  we  shall  have 

QXDB=RXDC. 


For,  by  applying  to  the  point  A  a  third  force  S,  equal 
and  directly  opposed  to  the  resultant  R,  the  three  forces 
p,  Q,  S,  will  be  in  equilibrium ;  consequently  the  force  P 
will  be  equal  and  directly  opposed  to  the  resultant  of 
the  two  forces  Q,  S.  Hence,  the  moments  of  the  two 
forces  Q,  S,  referred  to  the  point  D  in  the  direction  of 
their  resultant,  will  be  equal  (80) ;  hence  we  shall  have 


QXDB=SXDC; 


or,  since  S=R, 


QXDB=RXDC. 


MOMENTS. 


77 


we  have,  in  the  first  case, 


THEOREM. 

82.    When  the  directions 
of  the   two  forces   P,   Q, 
meet  in  the  same  point  A, 
the  moment  of  the  result 
ant  E  of  these  forces  re 
ferred  to  any  point  D,  ta 
ken  in  the  plane  of  these 
directions,  is  equal  to  the 
sum  or  difference  of  the 
moments  of  the  forces  P,  Q, 
referred  to  the  same  point,  ac 
cording  as  the  point  D  is  with 
out  or  within  the  angle  PAQ, 
formed  by  the  directions  of 
these  forces :  that  is  to  say,  if 
from  the  point  D  we  let  fall 
upon    these    directions,    and 
upon   that   of  the   resultant, 
the  perpendiculars  DB,  DC,  DE, 


and,  in  the  second  case, 


RXDE  =  QXDC— -PXDB. 

7* 


78  STATICS. 

DEMONSTRATION.  Draw  the  line  AD,  and  decompose 
the  force  P  into  two  others,  p,  p',  the  first  in  the  direc 
tion  AD,  and  the  second  in  the  direction  of  the  force  Q.  For 
this  purpose  (38),  represent  the  force  P  by  the  part  AF  of 
its  direction  ;  through  the  point  r  draw  the  lines  FG,  FH, 
respectively  parallel  to  AQ  and  AD  ;  and  the  two  com 
ponents  p,  p'  will  be  represented  by  the  sides  AG,  AH 
of  the  parallelogram  AGFH. 

The  point  D  being  upon  the  direction  of  the  compo 
nent  p,  the  moment  of  the  other  component  p',  referred 
to  this  point,  is  equal  to  the  moment  of  their  resultant 
P,  and  we  shall  have  (81) 


Moreover,  taking  the  two  forces  p,  p'  instead  of  the 
force  P,  the  resultant  R,  of  the  two  forces  P,  Q,  is  also 
the  resultant  of  the  three  forces  p,  p',  Q. 

This  being  done,  in  the  first  case,  the  two  forces  Q 
and  pf  (Fig.  28),  which  act  along  the  same  line  of  di 
rection,  are  equivalent  to  a  single  force  equal  to  their 
sum  Q+y  ;  thus  the  force  R  may  be  regarded  as  the  re 
sultant  of  the  two  forces  p  and  Q+p'  ;  hence  the  moment 
of  this  resultant  referred  to  the  point  D,  in  the  direction 
of  the  first  of  these  forces,  is  equal  to  the  moment  of 
the  second  (81)  ;  hence  we  shall  have 

RXDE=(Q+J/)  DC, 
or 

QXDC-f//XDC. 


MOMENTS. 


79 


Hence,  by  substituting  for  the  moment  p'xvc  its 
value,  we  have 


In  the  second  case,  the  two  forces  Q,  pf  (Fig.  29), 
which  are  in  the  same  line,  and  which  act  in  contrary 
directions,  are  equivalent  to  a  single  force  equal  to  their 
difference  Q—  p'  :  now,  the  moment  of  this  single  force, 
referred  to  the  point  D  in  the  direction  of  the  force  p, 
is  equal  to  the  moment  of  the  resultant  R  of  these  two 
forces  (81)  ;  and  we  have 


or 


RXDE=(Q—  p') 


RXDE  =  QXDC—    ' 


Hence,  substituting  for  the  moment  p'xvc  its  value, 
we  have 

RXDE  =  QXDC  —  PXDB. 


HemarJc  I. 

83.  This  theorem  (82)  of  Statics  is  a  consequence  of 
the  following  geometrical  proposition  : 

Fig.  28.   (a)  If  from  any  point  D,  taken  in  the 

plane  of  a  parallelogram  AFML,  we 
let  fall  the  perpendiculars  DB,  DC, 
DE,  upon  the  sides  and  the  diagonal 
of  this  parallelogram,  which  meet 
in  the  same  point  A,  the  product 


80 


STATICS. 


of  the  diagonal  AM  by  its  perpendicular  DE  is  equal  to 
the  sum  of  the  products  of  the  sides  multiplied  each  by 
its  perpendicular  DB  and  DC.  Among  the  known  demon 
strations  of  this  theorem,  the  following  is  one  of  the 
simplest :  The  triangles  ADF,  ADL,  ADM,  having  the  same 
base  AD,  are  to  each  other  as  their  altitudes  F/,  L/,  MWZ  ; 
but  we  have  Mm— F/+LZ;  for  drawing  LK  parallel  to 
AD,  we  have  Mm=MK-f  Kw=F/-f-L£;  hence  the  triangle 
ADM  is  equal  to  the  sum  of  the  triangles  ADF  and  ADL  : 
hence 

AMXDE=AFXDB  +  ALXDC. 


TtemarTc  II. 


Fig.  29. 


84.  If  we  suppose  the  line  AD  to  be  inflexible  and  the 
point  D  immoveable ;  when  this  point  is  placed  without 
the  angle  PAQ  (Fig.  28),  the  two  forces  P,  Q  tend  to  turn 
the  point  A  in  the  same  direction  around  the  point  D ; 
and,  on  the  contrary,  when  the  point  D  is  placed  within 
the  angle  PAQ  (Fig.  29),  the  two  forces  tend  to  turn  the 
point  A  in  opposite  directions. 


MOMENTS.  81 

Hence,  if  two  forces  be  directed  in  the  same  plane, 
the  moment  of  their  resultant,  referred  to  any  point 
taken  in  this  plane,  is  equal  to  the  sum  or  difference  of 
^heir  moments  referred  to  the  same  point,  according  as 
these  forces  tend  to  turn  their  point  of  application 
around  the  centre  of  moments,  either  in  the  same  or  in 
opposite  directions ;  and,  in  all  cases,  the  resultant  tends 
to  turn  its  point  of  application  in  the  same  direction  as 
that  of  the  two  forces  whose  moment  is  the  greater. 


THEOREM. 

85.  When  the  forces  P,  Q,  R,  S (Fig.  30),  di 
rected  in  the  same  plane,  are  applied  to  the  points 
a,  b,  c,  d,  connected  together  in  an  invariable  manner, 
tending  to  turn  these  points  in  the  same  direction  around 
another  point  D  taken  in  this  plane,  the  sum  of  the  mo 
ments  of  these  forces,  referred  to  the  point  D,  is  egual  to 
the  moment  of  their  resultant  referred  to  the  same  point. 

DEMONSTRATION.  Let  v  be  the  partial  resultant  of 
the  two  forces  P,  Q ;  x  that  of  the  two  forces  V,  R,  and 
consequently  of  the  three  forces  P,  Q,  R ;  Y  that  of  the 
two  forces  x,  S,  and  consequently  of  the  four  forces  P, 
Q,  R,  S  ;  and  so  on.  Then,  from  the  point  D  let  fall  upon 
the  directions  of  the  forces  and  upon  those  of  the  par 
tial  resultants  V,  x,  Y,  .  .  .  the  perpendiculars  DE,  DF, 
DG,  DH,  .  .  .  .  DI,  DK,  DL,  .  .  .  This  being  done,  the 
moment  of  the  resultant  v  is  equal  to  the  sum  of  the 
moments  of  its  components  P,  Q,  (82),  which  gives 

VXDI=PXDE-f  QXDF. 


In  like  manner  the  moment  of  the  resultant  X  is  equal 
to  the  sum  of  the  moments  of  its  components  v,  R,  and 
we  have 


or,  substituting  for  the  moment  VXDI  its  value, 

XXDK—  PXDE  +  QXDF  +  RXDG. 

Likewise,  we  have  YXDL=XXDK+SXDH,  or,  substi 
tuting  the  value  of  XXDK, 

YXDL=PXDE  +  QXDF  +  RXDG  +  SXDH  ; 


MOMENTS. 


83 


and  so  on,  whatever  may  be  the  number  of  forces. 
Hence  the  moment  of  each  resultant  is  equal  to  the  sum 
of  the  moments  of  all  its  components. 

COROLLARY  I. 

86.  If  the  forces  P,  Q,  R,  S  .  .  .  .  do  not  tend  to  turn 
their  points  of  application  in  the  same  direction  around 
the  centre  of  the  moments  D,  the  moment  of  their  re 
sultant  is  equal  to  the  excess  of  the  sum  of  the  moments 
of  the  forces  which  tend  to  turn  in  one  direction,  over 
the  sum  of  the  moments  of  those  which  tend  to  turn  in 
the  opposite  direction. 

Fig.  31 . 


Thus,  let  v  be  the  partial  resultant  of  all  the  forces 
p,  Q,  .  .  .  which  tend  to  turn  in  one  direction ;  and  let 
X  be  the  partial  resultant  of  all  the  forces  R,  S,  .  .  . 
which  tend  to  turn  in  the  opposite  direction ;  from  the 
point  D  let  fall  upon  the  directions  of  the  forces  and  upon 
those  of  the  two  resultants  v,  x,  the  perpendiculars  DE, 
DF,  .  .  .  DG,  DH,  .  .  .  DI,  DK ;  we  have  just  seen  (85), 
that  we  have 


84 


and 


STATICS. 


VXDI=PXDE  +  QXDF 


XXDK=RXDG  +  SXDH 


Finally,  let  Y  be  the  resultant  of  the  two  forces  v,  x, 
and  consequently  that  of  all  the  forces  P,  Q,  R,  s,  .  .  . 

This  being  established,  the  moment  of  the  resultant 
Y  referred  to  the  point  D  is  equal  to  the  difference  of 
the  moments  of  its  components  v,  x,  which  tend  to  turn 
in  opposite  directions  (84)  :  that  is  to  say,  by  letting  fall 
the  perpendicular  DL  upon  its  direction,  we  have 

YXDL=VXDI—  XXDK. 

Hence,  by  substituting  the  values  of  the  two  moments, 
we  have 


YXDL=PXDE  +  QXDF  .  .  .  — 


COROLLARY  II. 

8T.  If  the  directions 
of  the  forces  P,  Q,  R,  s, .  . . 
in  eluded  in  the  same  plane, 
are  parallel  to  each  other, 
the  perpendiculars  DE,  DF, 
DG,  DH,  .  .  .  DL  let  fall 
from  the  centre  of  mo 
ments  D  upon  these  directions  and  upon  that  of  the  re 
sultant  Y,  will  be  in  the  same  straight  line ;  and  the 


MOMENTS. 


85 


Fig.  33. 


preceding  proportion  -will  remain  just  the  same,  whether 
all  the  forces  act  in  the  same  direction,  as  in  Fig.  32,  or 
whether  they  act,  some  in  one  direction  and  others 
in  the  ODposite  direction,  as  in  Figs.  33  and  34.  Now, 
the  resultant  Y  of  all  these 
forces  is  equal  to  the  excess  of 
the  sum  of  those  which  act  in 
one  direction,  over  the  sum  of 
those  which  act  in  the  opposite 
direction  (29) ;  hence  the  dis 
tance  DL  of  the  centre  of  mo 
ments  from  the  direction  of 
the  resultant,  is  equal  to  the 
quotient  of  the  excess  of  the  sum 
of  the  moments  of  the  forces  which 
tend  to  turn  in  one  direction,  over 
the  sum  of  the  moments  of  those 
which  tend  to  turn  in  the  opposite 
direction,  divided  by  the  excess  of 
the  forces  which  act  in  one  direc 
tion,  over  the  sum  of  those  which 
act  in  the  contrary  direction  :  thus 
we  have 


Fig.  34. 


DL  = 


DL 


PXDE+QXDF  .  .  .— (RXDG+SXDH  . . .) 


(Fig.  32.) 


PXDE+QXDF  .  .  .—  (RXDG+SXDH  .  .  .) 

--  7  -  —  —  —  r  -  -, 
P  +  Q  .  .  .—  (R  +  S  .  .  .) 


DL_  PXDE+RXDG  .  .  .—  (QXDF+SXDH  .  .  .) 

p+s  .  .  .—  (Q+  R  .  .  .)  •'  *   ig'      *' 


86  STATICS. 

In  all  cases,  the  resultant  acts  in  the  direction  in 
which  the  sum  of  the  forces  is  the  greater ;  and  it  is 
placed,  with  reference  to  the  point  D,  on  the  side  on 
which  the  sum  of  the  moments  is  the  greater. 


8T 


CHAPTER  THIRD. 

ON    CENTRES    OF    GRAVITY. 

88.  THE  property  by  virtue  of  which  bodies,  left  to 
themselves,  fall  towards  the  earth  is  named  gravitation 
or  gravity. 

All  the  molecules,  of  which  bodies  are  composed,  have 
gravitation,  and  they  always  have  it ;  for  into  whatever 
number  of  parts  a  body  is  divided,  each  of  these  parts 
continually  gravitates,  and  falls  towards  the  earth  when 
left  to  itself. 

89.  The  effort  which  a  body  makes  to  fall,  when  it  is 
retained  or  supported  by  an  obstacle  which  opposes  its 
fall,  is  called  the  weight  of  the  body  ;  this  weight  may 
be  regarded  as  the  effect  of  a  force  which  is  constantly 
applied  to  the  body :  thus  we  are  accustomed  to  consider 
gravity  as  a  force. 

Gravity  is  not  a  force  rigorously  constant  for  the  same 
molecule ;  it  varies  according  to  the  different  positions 
which  this  molecule  has  relatively  to  the  sphere  of  the 
earth. 

1st.  When  the  distance  of  the  molecule  from  the  cen 
tre  of  the  earth  changes,  its  gravity  decreases  in  the 
same  ratio  as  the  square  of  this  distance  increases ;  be 
sides,  the  earth  not  being  perfectly  spherical,  and  the 
lines  drawn  from  its  centre~to  the  equator  being  greater 


CQ  STATICS. 

than  those  which  terminate  at  the  poles,  the  gravity  at 
the  surface  of  the  earth  is  greater  for  the  same  mole 
cule,  when  this  molecule  is  placed  at  the  poles,  than 
when  it  is  at  the  equator,  because  there  the  distance  of 
the  molecule  from  the  centre  of  the  earth  is  less. 

2d.  The  earth  turns  around  its  axis,  and  all  the  parts 
composing  it  perform  their  revolutions  in  the  same  time, 
that  is  to  say,  in  about  twenty-four  hours.  The  parts 
of  the  surface  near  the  equator  describe  greater  circum 
ferences  of  circles  than  those  described  by  the  parts 
near  the  poles ;  their  centrifugal  force,  which  likewise 
is  greater,  destroys  a  greater  part  of  the  effect  of  gravi 
tation,  and  is  a  new  cause  which  renders  this  latter  force 
less  at  the  equator  than  it  is  at  the  poles. 

Thus,  rigorously  speaking,  gravity  is  variable  for 
the  same  molecule,  'when  this  molecule  departs  from  or 
approaches  the  surface  of  the  earth,  and  when  it  departs 
from  or  approaches  the  equator :  but  the  distances  of  the 
positions  in  which  we  are  accustomed,  in  Statics,  to 
consider  the  same  molecule,  are  so  small  with  reference 
to  the  radius  of  the  earth,  that  the  effects  of  this  varia 
tion  are  absolutely  insensible ;  and  we  may  regard 
gravity  as  a  constant  force  for  the  same  molecule,  what 
ever  its  position  may  be. 

90.  The  straight  line  along  which  a  molecule,  aban 
doned  to  itself,  falls  to  the  earth,  and  which  is  evidently 
the  direction  of  gravitation,  is  named  a  vertical ;  this 
line  is  everywhere  perpendicular  to  the  surface  of  the 
earth,  or,  more  exactly,  to  the  surface  of  undisturbed 
water. 

91.  A  plane  is  said  to  be  horizontal  when  it  is  per 
pendicular  to  a  vertical. 


CENTRE    OF    GRAVITY.  89 

If  the  earth  were  perfectly  spherical  all  the  lines  of 
direction  of  gravitation  would  meet  in  the  same  point, 
which  would  be  the  centre :  but  the  earth  not  being  a  per 
fect  sphere,  the  lines  of  direction  of  gravitation  for  two 
different  molecules  may  not  be  in  the  same  plane ;  and 
when  they  are  in  the  same  plane,  they  meet  in  the  same 
point. 

However,  the  molecules  in  the  same  body,  and  those 
of  the  different  bodies  we  are  accustomed  to  consider  in 
Statics,  are  so  near  each  other  compared  with  their 
distances  from  the  centre  of  the  earth,  that  the  angle 
formed  by  the  directions  of  gravitation  for  any  two  of 
them  is  not  sensible,  and  we  may  regard  all  these  direc 
tions  as  parallel. 

92.  We  will  regard,  then,  all  the  molecules  of  heavy 
bodies  as  constantly  pushed  or  drawn  towards  the  earth 
by  forces  constant  for  each  of  them  ;  we  will  suppose 
that  these  forces  are  parallel,  and  act  in  the  same  di 
rection  ;  and,  consequently,  we  will  be  able  to  apply  to 
them  all  we  have  said  of  the  composition,  decomposition, 
and  equilibrium  of  parallel  forces. 

Now,  when  several  parallel  forces,  acting  in  the  same 
direction,  are  applied  to  points  invariably  connected  to 
gether,  we  have  seen :  1st,  that  these  forces  have  a  re 
sultant  equal  to  their  sum  (27);  2d,  that  the  direction 
of  this  resultant  is  that  of  the  components ;  3d,  that 
there  exists  a  centre  of  forces  through  which  this  re- 
sultant  always  passes,  even  though  the  forces,  without 
changing  in  intensity  and  without  ceasing  to  be  parallel,, 
should  change  in  direction  (30). 

Hence,  1st,  the  weights  of  all  the  molecules  of  a  solid 
body  have  a  resultant  which  constitutes  the  weight  of 


90  STATICS. 

the  body,  and  this  resultant  is  equal  to  the  sum  of  the 
weights  of  the  molecules ;  2d,  the  direction  of  this  re 
sultant,  or  of  the  weight  of  the  body,  is  always  parallel 
to  that  of  gravitation,  and  consequently  vertical;  3d, 
whatever  may  be  the  different  positions  given  to  this 
body,  the  directions  of  the  resultants  for  all  these  posi 
tions  meet  in  the  same  point ;  for  by  varying  the  position 
of  the  body,  the  intensity  of  the  forces,  which  act  upon 
the  molecules,  is  not  altered,  and  these  forces,  which 
only  change  in  direction  with  reference  to  the  bodies, 
do  not  cease  to  be  parallel  to  each  other. 

93.  The  point  through  which  the  direction  of  the 
weight  of  a  body  always  passes,  whatever  may  be  its 
position,  is  named  the  centre  of  gravity. 

94.  When  several  bodies  are  invariably  connected  to 
gether,  and  we  consider  their  assemblage  as  though  they 
made  but  one  and  the  same  body,  we  ordinarily  give  to 
this  assemblage  the  name  of  system. 

95.  Every  thing  that  has  just  been  said  of  a  single 
body  may  likewise  be  said  of  a  system  of  several  bodies  : 
that  is  to  say,  the  weight  of  the  system  is  equal  to  the 
sum  of  the  partial  weights  of  the  bodies  which  compose 
it ;  that  the  direction  of  this  weight  is  vertical,  and  that 
this  direction,  whatever  may  be  the  position  of  the  sys 
tem,   always  passes   through  the   same  definite  point, 
which  is  the  centre  of  gravity  of  the  system. 


COROLLARY  I. 

96.   We  may  always  regard  the  weight  of  a  body, 
or  system  of  several  bodies,  as  a  force  directed  verti- 


CENTRE   OF   GRAVITY.  91 

cally,  and  applied  to  the  centre  of  gravity  of  the  body 
or  system  :  for  this  weight,  which  is  the  resultant  of  the 
partial  weights  of  all  the  molecules  which  compose  the 
body  or  system,  may  be  considered  as  applied  to  any 
point  of  its  direction,  and  consequently  to  the  centre  of 
gravity,  which  is  always  upon  this  direction,  whatever 
may  be  otherwise  the  position  of  the  body  or  the  system. 


COROLLARY  II. 

97.  Hence,  we  will  produce  an  equilibrium  in  the 
action  which  gravitation  exerts  upon  all  the  molecules 
of  a  body  or  system  of  bodies,  by  applying  to  the  centre 
of  gravity  of  the  body  or  system,  a  single  force  whose 
direction  is  vertical,  equal  to  the  total  weight  of  the 
body  or  system,  and  which  acts  in  a  direction  oppo 
site  to  that  of  gravitation. 

Inversely,  when  a  single  force  produces  an  equilibrium 
in  the  weight  of  all  the  molecules  of  a  body  or  system 
of  bodies,  the  direction  of  this  force  will  be  vertical, 
and  it  will  pass  through  the  centre  of  gravity  of  the 
body  or  system. 

Thus,  when  a  body  AB,  suspended  by  a 
thread  ED  from  a  fixed  point  D,  is  in  equi 
librium,  and  the  action  of  gravitation  is  con 
sequently  destroyed  by  the  resistance  of  the 
thread,  the  direction  of  this  thread  will  be 
vertical,  and  its  prolongation  will  pass 
through  the  centre  of  gravity  c  of  the  body. 


92  STATICS. 

COROLLARY  III. 

98.  From  this,  we  deduce  a  simple  man 
ner  of  finding,  by  experiment,  the  cen 
tre  of  gravity  of  a  body  of  any  figure. 
Thus,  if  we  suspend  the  same  body  by  a 
thread,  successively  by  the  two  different 
points  E,  E',  and  conceive  the  two  directions 
of  the  thread  to  be  prolonged  into  the  interior  of  the 
body,  the  point  c,  in  which  these  two  directions  inter 
sect,  will  be  the  centre  of  gravity  required. 

Remark. 

99.  Since  the  partial  weights  of  the  bodies  which 
compose  a  system,  may  be  considered  as  parallel  forces, 
applied  to  the  partial  centres  of  gravity  of  these  bodies, 
it  follows,  that  when  we  know  the  weight  of  these  bodies, 
and  the  positions  of  their  partial  centres  of  gravity,  we 
can  find  the  position  of  the  centre  of  gravity,  by  the 
processes  which  have  been  given  to  find  the  centres  of 
parallel  forces,  either  by  means  of  the  principle  of  the 
composition  of  parallel  forces,  as  in  No.  28,  or  by  em 
ploying  the  consideration  of  moments,  as  in  Nos.  75, 
77,  78,  and  79 ;  we  will  soon  have  occasion  to  give  ex 
amples. 

To  find  the  centre  of  gravity  of  a  body  of  any  figure 
whatever,  conceive  the  body  to  be  divided  into  a  certain 
number  of  parts,  so  that  we  may  know  the  weight  of 
each  of  them,  and  the  position  of  its  partial  centre  of 
gravity ;  then,  by  finding  the  centre  of  gravity  of  the 
system  of  all  these  parts,  we  will  have  the  required  cen 
tre  of  gravity  of  the  body. 


CENTRE   OF   GRAVITY.  93 

But  when  the  parts  of  the  body  are  of  the  same  na 
ture  in  all  its  extent,  and  when  the  figure  of  the  body 
is  not  very  complicated,  we  may  often  find  its  centre  of 
gravity  by  simpler  considerations,  and  which  we  are 
about  to  employ  in  order  to  arrive  at  results  which  are 
used  very  frequently. 


LEMMA. 

Fig-  86.  100.    When  a  body  is  considered 

to  be  composed  of  parts  A,  A',  B,  B', 
c,  c',  .  .  .  which,  taken  two  and  two, 
are  equal  to  each  other,  and  so  placed 
that  the  middle  of  the  lines  AA', 
BB',  cc',  which  join  the  centres  of 
gravity  of  the  homologous  parts,  co 
incide  in  the  same  point  D,  this  point,  which  is  the 
centre  of  figure  of  the  body,  is  also  its  centre  of  gravity. 
For  the  point  D  is  the  centre  of  gravity  of  each  par 
tial  system  of  two  homologous  parts ;  hence,  it  is  also 
the  centre  of  gravity  of  their  general  system. 


COROLLARY. 

101.  By  considering  lines,  surfaces  and  solids,  as 
composed  of  parts  uniformly  heavy,  it  is  evident :  1st, 
that  the  centre  of  gravity  of  a  right  line  is  at  the  mid 
dle  of  its  length  ; 


94 


STATICS. 


2d.  The  centre  of  gravity  of  the  area, 
and  that  of  the  contour  of  a  parallelo 
gram  ABCD,  are  in  its  centre  of  figure : 
that  is  to  say,  at  the  point  of  intersec- 
tion  E,  of  its  two  diagonals  AC,  BD  ; 

D  C 

3d.  The  centre  of  gravity  of  the  area  of  a  circle,  and 
that  of  its  whole  circumference,  are  at  the  centre  of  the 
circle ; 

4th.  The  centre  of  gravity  of  the  whole  surface  of  a 
parallelopipedon,  and  that  of  its  solidity,  are  in  its  cen 
tre  of  figure :  that  is  to  say,  in  the  intersection  of  any 
two  of  its  four  diagonals,  or  in  the  middle  of  one  of 
them; 

5th.  The  centre  of  gravity  of  the  convex  surface  of 
a  right  or  oblique  cylinder,  and  that  of  its  solidity,  are 
in  the  middle  of  the  length  of  its  axis  ; 

6th.  The  centre  of  gravity  of  the  surface  of  a 
sphere,  and  that  of  its  solidity,  are  at  the  centre  of  the 
sphere. 


PROBLEM. 

102.   To  find  the  centre  of  gravity  of  the  area  of  any 
rectilinear  triangle  ABC. 

Fig.  38.  SOLUTION.  Having  drawn  through 

the  summit  A  of  one  of  the  angles, 
and  through  the  middle  D  of  the  op 
posite  side,  the  line  AD,  if  we  con 
ceive  the  area  of  the  triangle  to  be 
divided  into  an  infinite  number  of 
elements  by  lines  parallel  to  BC,  the 


CENTKE    OF    GRAVITY.  95 

centre  of  gravity  of  each  of  these  elements  will  be  in 
its  middle  (101),  and  consequently  upon  the  line  AD; 
hence  the  centre  of  gravity  of  their  system,  which  will 
be  that  of  the  area  of  the  triangle,  will  be  upon  this 
same  line  (30).  For  the  same  reason,  if  from  the  sum 
mit  B  of  another  angle,  and  through  the  middle  E  of 
the  opposite  side,  we  draw  a  line  DE,  this  second  line 
will  contain  the  centre  of  gravity :  hence  this  centre 
will  be  found  both  upon  the  line  AD,  and  upon  the  line 
BE  ;  hence  it  will  be  found  at  the  point  of  intersection 
F  of  these  two  lines. 


COROLLARY  I. 

103.  If  from  the  summit  A  of  one 
of  the  angles  of  the  triangle  ABC, 
and  through  the  middle  D  of  the  op 
posite  side,  we  draw  a  line  AD,  and 
divide  this  line  into  three  equal  parts, 
the  centre  of  gravity  F  of  the  area 
of  the  triangle  will  be  upon  this  line, 
at  the  distance  of  two-thirds  from  the  summit  of  the 
angles,  or  one-third  from  the  opposite  side. 

For,  if  we  draw  the  line  DE,  this  line  will  be  parallel 
to  AB,  because  the  sides  BC,  AC,  are  cut  proportionally 
in  D  and  E ;  and  the  triangles  ABF,  DEF  will  be  similar, 
because  their  corresponding  angles  will  be  equal ;  hence 
we  shall  have 

AF  :  FD  : :  AB  :  DE. 
But  the  similar  triangles  ABC,  EDC,  give 


96  STATICS. 

AB  :  DE  :  :  BC  :  DC,  or  :  :  2  :  1  (102). 
Hence  we  shall  have 

AF  :  FD  : :  2  :  1,  or  AF=2FD, 


hence, 


FD=JAD,  and  AF=§  AD. 


COROLLARY  II. 

104.  If  in  the  plane  of  a 
rectilinear  triangle  ABC  we  draw 
any  line  GI,  the  perpendicular 
let  fall  from  the  centre  of  gravi 
ty  F  of  the  area  of  the  triangle 
upon  GI,  will  be  equal  to  one- 
third  of  the  sum  of  the  perpen 
diculars  AG+CH+BI  let  fall 
from  the  summits  of  the  angles 
upon  the  same  line. 

Thus,  through  the  summit  A  of  one  of  the  angles 
draw  the  line  AM  parallel  to  GI,  which  will  cut  in  K,  L, 
M,  the  perpendiculars  let  fall  from  the  other  points ; 
through  the  point  A,  and  through  the  centre  of  gravity 
F,  draw  the  line  AF,  whose  prolongation  will  bisect  the 
opposite  side  at  the  point  D ;  finally,  through  the  point 
D,  draw  DN  perpendicular  to  AM  :  this  being  done,  we 
shall  have 

CK+BM 


CENTRE    OF    GRAVITY.  97 

and  the  similar  triangles  AFL,  ADN  will  give 

FL  :  DN  : :  AF  :  AD,  or  : :  2  :  3  (103). 
Hence, 

CK+BM 

FL=fDN  = g . 

But  the  lines  AG,  KH,  LO,  MI,  being   equal  to  each 
other,  we  shall  have 


LO= 


AG-f-KH-f-MI 

~3         ' 


Hence,  by  adding  this  equation  to  the  preceding,  we 
shall  have 


AG  +  CK  +  KH  +  BM-fMI 


that  is  to  say, 


FO' 


AG-fCH  +  BI 

:          3         !* 


From  this  we  deduce 
another  manner  of  finding 
the  centre  of  gravity  of 
the  area  of  a  rectilinear 
triangle  ABC.  Having 
drawn  at  pleasure,  in  the 
plane  of  the  triangle,  two 
lines  GI,  GP,  and  having 
found  the  distances  FO, 
FR,  of  the  centre  of  gravity  from  each  of  these  lines, 

9 


98 


STATICS. 


if  we  draw  the  line  RV  parallel  to  GI  and  at  the  distance 
FO,  this  line  will  contain  the  required  centre  of  gravity; 
in  like  manner  if  we  draw  xo  parallel  to  PG  and  at  the 
distance  FR,  this  second  line  will  contain  the  centre  of 
gravity ;  hence  this  centre  will  be  found  upon  the  two 
lines  RV,  xo ;  hence  it  will  be  at  their  point  of  inter 
section  F. 


Fig.  41. 


PROBLEM. 

105.   To  find  the  centre  of  gravity  of  the  area  of  a 
rectilinear  polygon  ABODE  of  any  number  of  sides. 

FIRST  SOLUTION,  by  the  process  of  the  composition 
of  parallel  forces. 

Divide  the  area  of  the  polygon 
into  triangles  by  the  diagonals 
AC,  AD,  .  .  .  drawn  from  the  sum 
mit  of  the  same  angle  A,  and  de 
termine  (102,  or  103,  or  104), 
the  partial  centres  of  gravity  F, 
G,  H  of  the  areas  of  these  trian 
gles  ;  then  considering  these  tri 
angles  as  weights  proportional 
to  their  areas  and  applied  to  their  centres  of  gravity, 
join  the  centres  of  gravity  of  the  first  two  triangles  ABC, 
CAD  by  a  line  FG,  and  find  upon  this  line  the  centre  of 
gravity  I  of  the  system  of  the  two  triangles,  or  of  the 
quadrilateral  ABCD,  by  dividing  the  line  FG  into  two 
parts  reciprocally  proportional  to  the  areas  of  the  two 


CENTRE    OF    GRAVITY.  99 

triangles  (18,)  which  may  be  done  by  the  following  pro 
portion  (25) : 

quadrilateral  ABCD  :  triangle  CAD  :  :  FG  :  Fi. 

Through  the  point  I,  and  through  the  centre  of  gravity 
H  of  the  next  triangle,  draw  the  line  Hi,  upon  which 
find  the  centre  of  gravity  K  of  the  system  of  the  first 
three  triangles,  by  dividing  this  line  into  two  parts  re 
ciprocally  proportional  to  the  areas  of  the  quadrilateral 
ABCD  and  of  the  triangle  DAE,  which  may  be  done  by 
the  following  proportion : 

pentagon  ABCD  :  triangle  DAE  : :  in  :  IK. 

+ 

By  thus  continuing,  whatever  may  be  the  number  of 
triangles,  we  will  find  the  centre  of  gravity  of  their  sys 
tem,  and  this  centre  will  be  that  of  the  area  of  the  pro 
posed  polygon. 

SECOND  SOLUTION,  taken  from  the  consideration  of 
moments. 

Having  divided  the  area 
of   the  polygon,  as  in  the 
preceding  solution,  and  de 
termined  the  partial  centres 
of  gravity  F,  G,  H,  .  .  of  all 
the  triangles,  draw  at  plea 
sure  in  the  plane  of  the  poly 
gon  two  lines  LM,  LN,  upon 
.    which  let  fall  perpendicu 
lars  from  all  the  centres  of  gravity  F,  G,  H,  .  . ;  con 
sider  these  lines  as  the  intersection  of  two  planes  paral- 


100  STATICS. 

lei  to  the  direction  of  gravitation.  This  being  done,  the 
distance  of  the  centre  of  gravity  of  the  polygon,  or  of 
the  system  of  all  the  triangles,  from  each  of  the  lines 
LM,  LN,  will  be  equal  to  the  sum  of  the  moments  of  the 
triangles  referred  to  each  plane,  divided  by  the  sum  of 
their  areas  (77) :  thus  the  distance  of  this  centre  from 
the  line  LM  will  be 

ABC  X  F/drCAD  X  GU/drDAE  X  H  Jl 
ABCDE 

and  its  distance  from  the  line  LN  will  be 

ABC  X  F/±CAD  X  Gf/'itDAE  X  H// 
ABCDE 

Hence,  by  drawing  a  line  parallel  to  LM,  and  at  a 
distance  equal  to  the  first  of  these  two  distances,  this 
line  will  contain  the  centre  of  gravity  of  the  polygon ; 
likewise  if  we  draw  a  line  parallel  to  LN,  at  a  distance 
equal  to  the  second  of  these  distances,  this  line  will  con 
tain  the  centre  of  gravity ;  hence  the  intersection  of 
these  two  lines  will  be  the  required  centre  of  gravity. 

106.  If  the  centres  of  gravity  F,  G,  II,  of  the  trian 
gles  which  compose  the  area  of  the  polygon,  were  not 
all  placed  on  the  same  side  with  reference  to  each  of  the 
lines  LM,  LN,  in  order  to  find  the  distance  of  the  centre 
of  gravity  K  of  the  polygon  from  each  of  these  lines, 
it  would  be  necessary  to  subtract  the  moments  of  the 
triangles,  whose  centres  of  gravity  are  situated  on  the 
other  side  of  this  line,  instead  of  adding  them  (77). 


CENTRE    OF    GRAVITY. 


101 


PROBLEM. 

107.   To  find  the  centre  of  gravity  of  the  contour  of  a 
polygon  ABCDE,  of  any  number  of  sides. 

FIRST  SOLUTION,  by  the  process  of  the  composition  of 
parallel  forces. 

Fi9-  43-  Bisect  each  of  the  sides 

of  the  polygon  at  the 
points  F,  G,  H,  i,  K,  which 
will  be  the  partial  cen 
tres  of  gravity  of  these 
sides  (101).  Then,  con 
sidering  all  the  sides  as 
weights  proportional  to 

J"  *•' y'  *'     *  their    lengths,    find    the 

centre  of  gravity  of  the  system  o  of  any  two  of  them, 
as  AB,  EC,  by  joining  their  centres  of  gravity  by  the  line 
FG,  and  dividing  this  line  into  two  parts  reciprocally 
proportional  to  these  sides,  which  may  be  done  by  the 
following  proportion  (22) : 

AB+BC  :  BC  : :  FG  :  FO. 

The  point  0  being  found,  draw  through  this  point,  and 
through  the  middle  H  of  the  next  side,  the  line  OH,  upon 
which  find  the  centre  of  gravity  P  of  the  system  of  the 
three  sides,  by  dividing  this  line  into  two  parts  recipro 
cally  proportional  to  the  side  CD  and  the  sum  of  the  first 
two,  AB,  BC  ;  which  may  be  done  by  the  proportion, 


AB+BC+ CD  :  CD  : :  on  :  OP. 

9* 


102  STATICS. 

In  like  manner,  drawing  the  line  PI,  find  the  centre 
of  gravity  Q  of  the  system  of  the  four  sides  AB,  BC,  CD, 
DE  by  the  proportion, 

AB-f  BC-fCD-f  DE  I  DE  :  :  PI  :  PQ. 

By  thus  continuing,  whatever  may  be  the  number  of 
sides  of  the  polygon,  find  the  centre  of  gravity  of  their 
system,  and  this  centre  will  be  that  of  the  contour  of  the 
polygon. 

SECOND  SOLUTION,  taken  from  the  consideration  of 
moments. 

Having  bisected  each  side  of  the  polygon,  draw  at 
pleasure  the  two  lines  LM,  LN,  upon  each  of  which  let 
fall  perpendiculars  from  the  middle  of  all  the  sides. 
This  being  done,  the  distance  of  the  centre  of  gravity 
K  of  the  system  of  all  the  sides  referred  to  each  of  the 
planes,  whose  lines  of  intersection  are  LM,  LN,  will  be 
equal  to  the  sum  of  the  moments  of  the  sides  referred 
to  this  plane,  divided  by  the  sum  of  the  sides  (77); 
thus,  the  distance  of  this  centre  from  the  line  LM  will  be 

AB  X  F/+  BC  X  Qff  +  CD  X  Ilh  +  DE  X  I/+  EA  X  K& 

AB  +  BC  +  CD-f  DE  +  EA 

and  its  distance  from  the  line  LN  will  be 


AB  +  BC  +  CD-f  DE  +  EA 

Hence,  drawing  a  line  parallel  to  LM,  at  a  distance 
equal  to  the  first  of  these  distances,  then  another  line 
parallel  to  LN,  at  a  distance  equal  to  the  second  of  these 


CENTRE    OF    GRAVITY. 


103 


distances,  the  point  of  intersection  of  these  two  lines 
will  be  the  centre  of  gravity  n  of  the  contour  of  the 
polygon. 

Remark. 

108.  If  the  middle  points  F,  G,  H,  I,  K  (Fig.  43), 
of  the  sides  of  the  polygon,  were  placed  on  opposite 
sides  of  the  lines  LM,  LN  ;  in  order  to  find  the  distance 
of  the  centre  of  gravity  R  from  each  of  these  lines,  it 
would  be  necessary  to  subtract  the  moments  of  the  sides 
whose  middle  points  were  situated  on  the  contrary  side, 
instead  of  adding  them  (77). 


PROBLEM. 

109.    To  find  the  centre  of  gravity  of  the  solidity  of 
any  triangular  pyramid  ABCD. 

SOLUTION.  Determine  the  centre 
of  gravity  F  of  the  area  of  one  of 
the  faces  BCD  of  the  pyramid  (103), 
by  drawing  through  the  summit  D  of 
one  of  the  angles  of  this  face,  and 
through  the  middle  E  of  the  oppo 
site  side  BC,  a  line  DE  ;  and  take 
upon  this  line  a  point  F  two-thirds 
of  the  distance  from  the  summit 
of  the  angle  or  one-third  from  the 
base ;  then  draw  the  line  AF.  This  being  done,  if  we 
conceive  the  pyramid  to  be  divided  into  an  infinite  num 
ber  of  sections  by  planes  parallel  to  the  face  DCB,  all 


104  STATICS. 

these  sections  will  be  similar  to  this  face,  and  they  will 
be  met  by  the  line  AF  in  points,  which,  being  situated 
upon  each  of  them  in  the  same  manner  as  the  point  F 
is  in  the  face  BCD,  will  be  the  partial  centres  of  gravity 
of  these  sections ;  hence  the  centre  of  gravity  of  their 
system,  which  will  be  that  of  the  solidity  of  the  pyra 
mid,  will  be  upon  the  line  AF  (30). 

For  the  same  reason,  having  determined  the  centre 
of  gravity  a  of  the  area  of  another  face  ABC,  which  is 
done  by  drawing  the  line  AE,  and  taking  upon  this  line 
the  part  EG=J-  AE  ;  if  through  this  point,  and  through 
the  summit  D  of  the  opposite  angle  of  the  pyramid,  the 
line  DG  be  drawn,  this  line  will  also  contain  the  centre 
of  gravity  of  the  solidity  of  the  pyramid. 

Hence  the  lines  AF,  Da,  both  containing  the  centre 
of  gravity  of  the  pyramid,  will  necessarily  intersect  in 
a  certain  point  H  ;  and  the  point  of  intersection  of  these 
two  lines  will  be  the  centre  of  gravity  required. 

Remark  I. 

110.  It  might  be  demonstrated,  independently  of  the 
consideration  of  the  centre  of  gravity  of  the  pyramid, 
that  the  lines  AF,  DG  necessarily  intersect  in  one  point ; 
for  these  lines  are  in  the  same  plane,  which  is  that  of 
the  triangle  ADE. 

Remark  II. 

111.  Any  one  of  the  six  edges  of  a  triangular  pyra 
mid  being  cut  by  four  others ;  the  fifth  which  does  not 
meet  it  is  called  its   opposite :  if  we  join  the  middle 
point  of  one  of  the  six  edges  with  that  of  its  opposite 


CENTRE    OF    GRAVITY. 


105 


by  a  line,  it  may  be  demonstrated  that  the  middle  of 
this  line  is  the  centre  of  gravity  of  the  pyramid.  (See 
the  Correspondence  of  the  Polytechnic  School,  tome  II, 
page  1.) 

COROLLARY  I. 

112.  If  from  the  summits  of  one  of 
the  angles  of  a  triangular  pyramid, 
and  through  the  centre  of  gravity  F 
of  the  area  of  the  opposite  face  BCD, 
a  line  AF  be  drawn,  the  centre  of 
gravity  H  of  the  solidity  of  the  pyra 
mid  will  be  upon  this  line,  and  at 
one-fourth  of  the  distance  from  the 
face,  or  at  three-fourths  of  the  dis 
tance  from  the  summit  of  the  angle. 
Draw  the  line  GF,  which  will  be  parallel  to  AD,  be 
cause  the  lines  EA,  ED,  are  cut  proportionally  in  G,  F ; 
the  triangles  AHD,  FHG,  whose  corresponding  angles  are 
equal,  will  be  similar,  and  will  give 

AH  :  IIF  : :  AD  :  GF. 
But  the  similar  triangles  AED,  GEF,  give 

AD  :  GF  : :  ED  :  EF,  or  : :  3  :  1  (103) ; 
hence,  we  shall  have 

AH  :  HF  : :  3  :  1 ; 


that  is  to  say,  AH=3HF,  and  consequently  HF=|AF,  and 
AH=?AF. 


106  STATICS. 


COROLLARY  II. 

113.  It  might  be  demonstrated,  in  a  manner  analo 
gous  to  that  of  No.  104,  that  the  distance  of  the  centre 
of  gravity  of  the  solidity  of  a  triangular  pyramid  from 
any  plane,  is  equal  to  the  fourth  of  the  sum  of  the  dis 
tances  of  the  summits  of  the  four  angles  of  the  pyra 
mid  from  the  same  plane. 


COROLLARY  III. 


Fig.  45.  114.   The  centre  of  gravity  0  of  the 

solidity  of  a  pyramid  ABCDEF  with 
any  base  is  upon  the  line  AG,  drawn 
from  the  summit  A  to  the  centre  of 
gravity  G  of  the  area  of  the  base,  and 
at  a  distance  of  one-fourth  of  this 
line  from  the  base  or  three-fourths 
from  the  summit. 

Conceive  the  pyramid  to  be  divi 
ded  into  an  infinite  number  of  sec 
tions  by  planes  parallel  to  the  base  : 
all  these  sections  will  be  similar  to  the  base,  and  the 
point  where  each  of  them  is  intersected  by  the  line  AG 
will  be  situated  upon  this  section  in  the  same  manner  as 
the  point  G  is  upon  the  base  ;  consequently  this  point 
will  be  the  centre  of  gravity  of  the  section  :  hence  the 
centres  of  gravity  of  all  the  sections  will  be  upon  the 
line  AG  ;  hence  the  centre  of  gravity  of  their  system, 


CENTKE    OP    GRAVITY.  107 

which  is  that  of  the  solidity  of  the  pyramid,  will  be 
also  upon  this  line  (30). 

Moreover,  let  the  base  be  divided  into  triangles  by 
the  diagonals  BE,  BD,  and  conceive  that  through  these 
diagonals  and  through  the  summit  A,  the  planes  ABE, 
ABD  be  drawn,  which  will  divide  the  proposed  pyra 
mid  into  as  many  triangular  pyramids  as  there  are  tri 
angles  in  the  base ;  then  through  the  centres  of  gravity 
H,  I,  K  of  the  triangular  bases,  draw  the  lines  AH,  AI, 
AK ;  finally,  let  the  points  L,  M,  N  be  taken  upon  these 
lines,  upon  each  of  them  at  the  distance  of  one-fourth 
of  its  length  from  the  base ;  these  points  will  be  the 
centres  of  gravity  of  the  triangular  pyramids  (112). 
This  being  done,  the  points  L,  M,  N,  which  will  divide 
proportionally  the  lines  AH,  AI,  AK,  drawn  from  the  sum 
mit  of  the  pyramid  upon  the  base,  will  be  in  the  same 
plane  parallel  to  the  base ;  hence  the  centre  of  gravity 
of  the  system  of  triangular  pyramids, — that  is  to  say, 
the  centre  of  gravity  of  the  solidity  of  the  proposed 
pyramid, — will  be  in  this  same  plane  ;  hence  the  centre 
of  gravity,  being  found  both  in  this  plane  and  in  the 
line  AG,  will  be  at  the  point  of  their  intersection  o. 

Now,  the  line  AG  will  be  cut  by  the  plane  LMN  in  parts 
proportional  to  the  divisions  of  the  lines  AH,  AI,  AK  ; 
hence  the  centre  of  gravity  o  of  the  solidity  of  the 
pyramid  will  be  placed  upon  AG,  at  one-fourth  of  this 
line  from  the  base,  or  three-fourths  from  the  summit. 


108 


STATICS. 

COROLLARY  IV. 


115.  The  centre  of  gravity  of  the  solidity  of  a  cone 
of  any  base  is  upon  the  line  drawn  from  the  summit  to 
the  centre  of  gravity  of  the  base,  and  at  the  distance 
of  one-fourth  of  this  line  from  the  base,  or  at  three- 
fourths  from  the  summit ;  for  this  solid  may  be  consid 
ered  a  pyramid  whose  base  has  an  infinite  number  of 
sides. 


PROBLEM. 

116.  To  find  the  centre  of  gravity  of  the  area  of  a 
section  made  in  the  hull  of  a  vessel  ly  a  horizontal 
plane. 

SOLUTION.  Let  CEuhec  be 
the  proposed  section,  AB  the 
line  of  intersection  of  the 
plane  of  this  section  with  the 
vertical  plane  drawn  through 
the  keel  of  the  vessel.  It  is 
evident,  since  the  whole  section  is  symmetrical  on  each 
side  of  the  line  AB,  the  required  centre'  of  gravity  K 
will  be  upon  this  line ;  thus,  to  construct  this  point,  it 
will  suffice  to  know  its  distance  AK  from  a  line  Cc,  drawn 
through  a  given  point  perpendicular  to  AB. 

For  this  purpose,  let  the  line  AB  be  divided  by  the 
perpendiculars  or  ordinates  vd,  EC,  F/j  .  .  .  into  a  suffi 
ciently  great  number  of  equal  parts,  so  that  the  arcs  CD, 
DE,  EF,  ....  included  between  two  adjoining  perpendi- 


CENTRE    OF    GRAVITY.  109 

culars,  may  be  regarded  as  right  lines,  which  will  divide 
the  area  of  the  section  into  trapeziums ;  then  let  each 
of  these  trapeziums  be  divided  into  triangles,  by  means 
of  the  diagonals  cd,  DC,  E/,  .  .  .  .  This  being  done,  if 
we  take  the  sum  of  the  moments  of  all  the  triangles  re 
ferred  to  the  vertical  plane  passing  through  the  line  cc9 
and  divide  this  sum  by  the  sum  of  the  areas  of  the  tri 
angles,  the  quotient  will  be  the  distance  required  AK 
(77).  Now,  each  triangle  may  be  considered  as  having 
for  base  one  of  the  perpendiculars,  and  for  height  the 
common  distance  from  each  other  of  two  consecutive 
perpendiculars ;  hence  the  area  of  each  triangle  will  be 
equal  to  the  half  of  the  product  of  the  ordinate  which 
serves  for  base,  multiplied  by  the  common  distance. 
For  example,  the  area  of  the  triangle  DEe  will  be  equal 
to  the  half  of  the  product  EeXLM  ;  that  of  the  triangle 
vde  will  be  the  half  of  DC?XLM,  and  so  of  the  others. 
Moreover,  the  distance  of  the  centre  of  gravity  of  each 
of  the  triangles  from  the  plane  cc,  will  be  equal  to  one- 
third  of  the  sum  of  the  distances  of  the  summits  of  its 
three  angles  from  the  same  plane  (104) :  for  example, 
the  distance  of  the  centre  of  gravity  of  the  triangle 
DEe  from  the  plane  cc  will  be  one-third  of  AL+AM+AM, 
and  so  of  the  others. 

Hence  it  will  be  easy  to  have  the  sum  of  the  areas  of 
all  the  triangles,  and  the  sum  of  the  moments  of  these 
areas  referred  to  the  plane  cc ;  and  by  dividing  the 
second  of  these  two  sums  by  the  first,  we  shall  have  the 
required  distance  of  the  centre  of  gravity  K  from  the 
line  cc. 

The  preceding  solution  is  not  rigorous,  because  the 

parts  CD,  DE,  .  .  .,  cd,  de,  .  .  .,  of  the  sides  of  the  sec- 

10 


110  STATICS. 

tion  arc  not  right  lines,  as  we  have  supposed ;  but  it  is 
evident  that  the  result  will  approach  exactness  as  much 
more  as  these  parts  are  smaller :  that  is  to  say,  as  the 
number  of  perpendiculars  are  greater. 

117.  The  operation  just  described  is  susceptible  of 
some  reduction.  Thus,  according  to  the  preceding,  the 
area  of  the  triangle 


cc 


That  of  CDcZ=ALX, 


That  of  vde=ALX~, 
2 


That  of  DEe=ALX— , 

JH 


That  of  E£>/=ALX??, 
A 


That  of  EF/=ALX^; 
a 


and  so  on  with  the  others.  By  adding  all  these  pro 
ducts  together,  we  see  that  their  sum  is  equal  to  the 
product  of  the  common  factor  AL,  multiplied  by  half  the 
sum  of  the  two  extreme  perpendiculars  and  the  sum  of 
all  the  others. 


CENTRE    OF   GRAVITY.  Ill 

As  to  the  moments  of  these  triangles  referred  to  the 
plane  C£,  we  have 

That  of  ccd=ALX  —  x-- 


mi  c  7  D^      2AL 

That  of  CD<i=ALX—  x—  ^-, 
2        o 


mi  T>d     4AL 

That  of  Ddpe=ALX-6-x-^-, 


That  Of  DE£=ALX  —  X 


2      3 
That  of  Eef=ALX-r 


That  of  EF/=ALX-X-^-, 


and  so  on  ;  in  which  we  see  that  the  number  which  mul 
tiplies  AL,  in  the  moment  of  the  last  triangle,  is  always 
equal  to  three  times  the  number  of  the  intervals  minus 
unity  ;  or,  which  is  the  same  thing,  to  three  times  the 
number  of  the  last  perpendicular,  less  4.  By  adding 
together  all  these  moments,  we  find  their  sum  equal  to 
the  product  of  the  common  factor  ALXAL,  multiplied  by 
the  sum  composed  of  one-sixth  of  the  first  perpendicu 
lar,  one  sixth  of  the  last,  multiplied  by  three  times  the 
number  of  perpendiculars  less  4,  then  of  the  second 
perpendicular,  double  the  third,  three  times  the  fourth, 
.  .  .  and  so  on. 


112  STATICS. 

Now  the  sum  of  the  moments  and  that  of  the  areas 
having  the  common  factor  AL,  their  quotient  will  also  be 
the  same  if  we  suppose  this  factor  to  be  in  both  terms 
of  the  division ;  hence,  to  obtain  the  distance  of  the 
centre  of  gravity  K  from  one  of  the  extreme  ordinates 
c<?,  it  is  necessary ',  1st,  to  take  one-sixth  of  the  first  or- 
dinate  Co  ;  one-sixth  of  the  last  nh,  multiplied  by  three 
times  the  number  of  ordinates.,  less  4  ;  then  the  second 
ordinate,  double  the  third.,  three  times  the  fourth,  .  .  . 
and  so  on ;  which  will  give  the  first  sum :  2d,  to  the 
half  of  the  two  extreme  ordinates  add  all  the  interme 
diate  ordinates  ;  this  will  give  the  second  sum :  3d,  di 
vide  the  first  of  these  two  sums  by  the  second,  and  multi 
ply  the  quotient  by  the  common  interval  of  the  ordinates. 


PROBLEM. 

118.  To  find  the  centre  of  gravity  of  the  volume  of 
the  submerged  part  of  the  hull  of  a  vessel. 

SOLUTION.  We  will  suppose  that  the  vessel,  being 
afloat,  has  its  keel  horizontal,  and  that  the  vertical  plane 
drawn  through  the  keel  divides  the  volume  of  the  hull 
into  two  perfectly  symmetrical  parts.  This  being  done, 
the  centre  of  gravity  of  the  part  submerged  will  be  in 
this  plane,  and  the  question  will  be  reduced  to  find  the 
distances  of  this  point  from  two  lines  of  known  position 
in  the  vertical  plane. 


CENTRE    OF   GRAVITY. 

Fig.  47. 


113 


B  I — 


~~x_                 _^-  —  7i 

J 

^ 

E'( 

c       / 

\ 

P 

/ 

3  

-~^^f 

/ 

'           M 

\ 

N 

/ 

Let  ABCDF  be  the  section  of  a  vessel  through  the  ver 
tical  plane  CD,  its  keel,  and  conceive  the  plane  of  flota 
tion,  or  the  section  made  in  the  vessel  at  the  level  of  the 
water,  to  be  represented  by  the  line  Rb  parallel  to  the 
keel.  Let  the  interval  of  the  two  lines  B&,  CD,  be  divi 
ded  into  a  certain  number  of  equal  parts  BE',  B'B",  B"B'" 
.  .  .  .,  and  through  each  point  of  division  suppose  there 

are  horizontal  sections  represented  by  B'5',  B"^" 

In  like  manner  let  the  line  B&,  from  the  point  B  of  the 
stern-post,  be  divided  into  equal  parts  BF,  FF',  F'F".  . . . ; 
and  through  each  point  of  division  imagine  vertical 
planes  to  be  arranged  perpendicular  to  the  keel,  and 
represented  by  the  lines  BB'",  F/,  F^.  .  .  . ;  the  sub 
merged  part  of  the  hull  will  be  divided  into  rectangular 
prisms,  whose  sides  will  be  perpendicular  to  the  vertical 
plane  drawn  through  the  keel,  and  which  will  be  termi 
nated  on  both  sides  at  the  surface  of  the  vessel.  (It  is 
necessary  that  the  divisions  of  the  lines  BB'",  Eb  should 
be  so  small  that  the  part  of  the  surface  of  the  vessel, 
which  terminates  each  prism,  may  be  regarded  as  a 
plane).  Finally,  let  each  rectangular  prism,  represented 
by  its  base  LMNP,  be  divided  into  two  triangular  prisms, 
by  a  diagonal  plane,  represented  by  MP. 

This  being  done,  1st,  each  triangular  prism  will  always 
be  divided  into  three  pyramids  of  the  same  base  as  the 


10* 


114  STATICS. 

prism  (Legendre's  Geometry,  Book  VI.),  and  each  of 
which  will  have  for  height  one  of  the  sides  of  the  prism  : 
hence,  if,  by  actual  measurement  of  the  vessel,  we  get 
the  length  of  all  the  sides,  it  will  be  easy  to  find  the 
solidity  of  each  pyramid,  by  multiplying  the  area  of  the 
common  base,  LMP,  by  one-third  of  the  side,  which 
measures  the  height  of  the  pyramid  ;  and  by  taking 
the  sum  of  all  these  solidities,  we  will  obtain  that  of  the 
submerged  part  of  the  hull.  2d.  The  moment  of  a  tri 
angular  pyramid  referred  to  a  plane,  being  equal  to  the 
product  of  the  solidity  of  the  pyramid,  multiplied  by 
one-fourth  of  the  sum  of  the  distances  of  the  summits 
of  its  four  angles  from  this  plane  (113),  it  will  be  easy 
to  find  the  moment  of  each  pyramid  referred  to  the  ver 
tical  plane  BB'"  or  to  the  horizontal  plane  CD  ;  because 
the  distances  of  the  summits  of  these  angles  from  each 
of  these  planes  are  known ;  and  by  taking  the  sum  of 
all  these  moments,  we  shall  have  the  moment  of  the 
submerged  part  of  the  hull. 

This  being  done,  the  quotient  of  the  sum  of  the  mo 
ments  referred  to  the  vertical  plane  BB'",  divided  by  the 
sum  of  the  solidities,  will  be  the  distance  KX  of  the  re 
quired  centre  of  gravity  from  the  vertical  BB'" ;  in  like 
manner  the  quotient  of  the  sum  of  the  moments  re 
ferred  to  the  horizontal  plane  CD,  divided  by  the  sum  of 
the  solidities,  will  be  the  distance  KY  of  the  same  point 
from  the  keel.  We  shall  have,  therefore,  the  distances 
of  the  centre  of  gravity  from  two  lines  of  known  posi 
tion  in  the  vertical  plane  drawn  through  the  keel,  and 
consequently  the  position  of  this  point  will  be  deter 
mined. 


CENTRE   OF   GRAVITY.  115 

The  preceding  solution  is  not  rigorous ;  because  the 
surface  of  the  vessel  being  curved,  the  part  of  this  sur 
face  which  terminates  each  triangular  prism  cannot  be 
regarded  as  a  plane,  as  we  have  supposed ;  but  the  re 
sult  will  approach  exactness  as  much  more  as  the  num 
ber  of  divisions,  both  in  the  direction  of  the  height  of 
the  vessel,  and  in  that  of  its  length,  are  greater. 

119.  The  operation  just  described  is  susceptible  of 
some  reduction;  and  by  reasoning  as  in  No.  116  we  find, 
that  to  get  the  distance  KX  of  the  centre  of  gravity  of 
the  submerged  part  of  the  hull  from  the  vertical  plane 
BB'",  it  is  necessary,  1st,  for  each  horizontal  section,  to  take 
one-sixth  of  the  first  ordinate  which  is  in  the  plane  BB'", 
one-sixth  of  the  last,  multiplied  l>y  three  times  the  num 
ber  of  ordinates  contained  in  the  section,  less  4  ;  then  the 
second  ordinate,  double  the  third,  three  times  the  fourth, 
.  .  .,  which  will  form  a  partial  sum  for  each  section; 
then  add  together  the  half  of  the  first  of  these  sums,  the 
half  of  the  last,  and  all  the  intermediate  ones,  which 
will  form  a  dividend  ;  2d,  to  one-fourth  of  the  four  or 
dinates  placed  at  the  angles  of  the  rectangle  EW's'",  add 
one-half  of  all  which  are  upon  the  contour  of  this  rect 
angle,  and  the  whole  of  all  those  in  the  interior,  which 
will  form  a  divisor ;  3d,  divide  the  dividend  by  the  di 
visor,  and  multiply  the  quotient  by  the  interval  BF  pa 
rallel  to  the  distance  required  KX. 

To  find  the  distance  KY  of  the  centre  of  gravity  from 
the  horizontal  plane  drawn  through  the  keel,  it  is  ne 
cessary  to  operate  upon  the  vertical  sections  as  upon 
the  horizontal  sections  in  the  preceding  case  :  that  is  to 
say,  1st,  for  each  vertical  section,  take  one-sixth  of  the 
lower  ordinates,  one-sixth  of  that  which  is  in  the  plane 


116  STATICS. 

of  flotation,  multiplied  by  three  times  the  number  of 
ordinates  of  the  section,  less  4 :  then  the  second  ordinate 
from  the  bottom,  double  the  third,  three  times  the  fourth, 
.  .  .  tuhich  ivill  form  for  each  section  a  partial  sum ; 
then  add  together  the  half  of  the  first  of  these  sums,  the 
half  of  the  last,  and  all  the  intermediate  ones,  which  will 
form  a  dividend ;  2d,  divide  this  dividend  by  the  same 
divisor  as  in  the  preceding  case,  and  multiply  the 
quotient  by  the  interval  BB  parallel  to  the  distance 
sought  KY. 


Remark. 

120.  In  the  preceding  problem,  the  only  object  is  to 
find  the  centre  of  gravity  of  the  volume  of  the  submerged 
part  of  the  hull,  or,  which  is  the  same  thing,  the  volume 
of  water  displaced  by  the  vessel.  But  if  it  were  re 
quired  to  find  the  centre  of  gravity  of  the  vessel  itself 
either  laden  or  unladen :  that  is  to  say,  to  find  the  dis 
tances  of  this  point  from  the  horizontal  plane  drawn 
through  the  keel,  and  from  the  vertical  plane  perpen 
dicular  to  the  keel,  it  would  be  necessary  to  take,  with 
reference  to  each  of  these  planes,  the  sum  of  the  mo 
ments  of  all  the  parts  which  compose  the  vessel  and  its 
load,  and  then  to  divide  each  of  these  sums  by  the  total 
weight  of  the  vessel  and  its  load ;  observing,  in  taking 
the  moments,  to  multiply,  not  the  volume,  but  the  weight 
of  each  part,  by  the  distance  of  the  partial  centre  of 
gravity  of  this  part  from  the  plane  to  which  the  moments 
are  referred ;  and  the  quotients  of  these  divisions  would 
be  the  distances  required. 


CENTRE    OF    GRAVITY.  117 

It  will  be  easy  to  find,  at  least  in  a  manner  sufficiently 
near,  the  partial  centre  of  gravity  of  each  of  the  parts 
of  the  vessel  and  its  load  ;  because  we  may  always  de 
compose  this  part  into  parallelopipedons,  cylinders,  py 
ramids,  or  other  solids  of  which  we  have  given  the 
means  of  finding  the  centres  of  gravity. 


118 


CHAPTER  FOURTH. 


ON   THE    EQUILIBRIUM    OF   MACHINES. 

121.  Every  instrument  intended  to  transmit  the  ac 
tion  of  a  determined  force,  to  a  point  which  is  not  found 
upon  its  direction,  so  that  this  force  may  move  a  body 
to  which  it  is  not  immediately  applied,  and  move  it  in 
a  direction  different  from  its  own,  is  called  a  machine. 

122.  In  general,  we  are  not  able  to  change  the  direc 
tion  of  a  force,  but  by  decomposing  this  force  into  two 
others ;  one  of  which  is  directed  toward  a  fixed  point, 
which  destroys  the  force  by  its  resistance,  and  the  other 
acts  in  a  new  direction :  this  latter  force,  which  is  the 
only  one  that  can  produce  any  effect,  is  always  a  com 
ponent  of  the  first ;  and  may  be  either  smaller  or  great 
er  than  it  is,  according  to  circumstances.    By  changing, 
in  this  manner,  the   directions  and  intensities  of  the 
forces,  we  may,  by  the  aid  of  a  machine  and  the  points 
of   support  which  it  presents,  produce  an  equilibrium 
between   two  unequal  forces  which    are   not   directly 
opposed. 

123.  The  force,  whose  direction  is  to  be  changed  by 
employing  a  machine,  is  ordinarily  named  a  power,  and 
the  term  resistance  is  applied  to  the  body  it  has  to 


MACHINES.  119 

move,  or  to  the  force  with  which  it  has  to  produce  an 
equilibrium  by  means  of  the  machine. 

124.  We  propose  here  to  find  only  the  ratios  wThich 
should  subsist  between  the  power  and  the  resistance  ap 
plied  to  the  same  machine,  in  order  that,  with  regard  to 
their  directions,  they  may  be  in  equilibrium.     We  will 
leave  friction  out  of  consideration :  that  is  to  say,  the 
difficulties  which  the  different  parts  of  the  machine  may 
experience  in    slipping  or   rolling   upon    each   other  ; 
and  we  will  suppose  that  cords,  when  they  enter  into 
the  composition  of  the  machine,  are  perfectly  flexible. 
Thus,  having  given  to  a  power  the  intensity  which  is 
proper  for  the  condition  of  equilibrium  in  this  supposi 
tion,  it  would  not  suffice  to  augment  it  by  a  small  quan 
tity  to  destroy  the  equilibrium  and  put  the  machine  in 
motion ;  but  we  must  first  apply  the  whole  quantity  re 
quired  to  overcome  the  obstacles  just  mentioned ;  and 
then  a  slight  increase  will  produce  motion. 

125.  Although  the  number  of  machines  is  very  great, 
we  may  regard  them  all  as  composed  of  three  simple 
machines,  which  are :  cords,  the  lever,  and  the  inclined 
plane.     We  will  content  ourselves  with  presenting  the 
theories  of  these  three  machines,  and  of  those  which 
are  immediately  derived  from  them ;  it  will  then  be 
easy,  by  simple  applications,  to  find  the  ratio  of  the 
power  to  the  resistance,  for  the  condition  of  equilibrium, 
in  every  machine,  however  complicated  it  may  be. 


120 


STATICS. 


AKTICLE  I, 

On  the  equilibrium  of  forces  which  act  upon  each  other 
by  means  of  cords. 

126.    We  will  suppose  that  the  cords   are  without 
weight ;  and  because  the  property  they  have  of  trans 
mitting  force,  is  independent  of  their  size,  we  will  sup 
pose  them  to  be  reduced  to  their  axes,  and  regard  them 
Fi9-  4a  as  straight,  flexible,  and  in- 

extensible  lines.  Taking 
this  for  granted,  let  us  con 
sider,  first,  the  case  of  equi 
librium  between  three  forces 
p,  Q,  R,  acting  upon  each 
other  by  means  of  three 
cords  united  together  by  a 
knot  A. 

1st.  The  three  forces  P, 
Q,  R,  cannot  be  in  equili 
brium,  unless  the  three  di 
rections,  and  consequently 
the  cords  by  means  of  which  they  transmit  their  actions, 
are  in  the  same  plane  (10). 

2d.  If  we  represent  any  two  of  these  forces,  for  ex 
ample,  the  forces  P,  Q,  by  the  parts  AC,  AD  of  their  di 
rections,  and  upon  these  lines  as  adjacent  sides  construct 
the  parallelogram  ACBD,  the  diagonal  AB  will  represent 
in  intensity  and  direction  the  resultant  of  these  two 
forces  (36) ;  now  the  three  forces  being  in  equilibrium, 
the  force  R  must  be  equal  and  directly  opposed  to  the 
resultant  of  the  two  others ;  hence  the  direction  of  the 


CORDS. 


121 


force  R  will  be  in  the  prolongation  of  BA,  and  its  in 
tensity  will  be  represented  by  this  diagonal :  thus  we 
shall  have 

p  :  Q  :  R  : :  AC  :  AD  :  AB  ; 

or  because  the  sides  AD,  BC  of  the  parallelogram  are 
equal  to  each  other,  the  three  forces  P,  Q,  R  will  be  to 
each  other  as  the  sides  of  the  triangle  ABC. 

127.  The  angles  of  the  triangle  ABC,  being  given  by 
the  directions  of  the  forces  P,  Q,  R,  and  the  magnitudes 
of  its  sides  being  proportional  to  the  intensities  of  these 
forces,  it  follows  that,  of  the  six  things  to  be  considered 
in  the  equilibrium  in  question,  namely,  the  directions 
of  the  forces  and  their  intensities,  any  three  being 
given,  we  can  find  the  three  others  in  all  cases  in  which, 
of  the  six  things  to  be  considered  in  the  triangle  ABC, 
namely,  the  angles  and  the  sides,  three  being  given,  we 
can  determine  the  three  others. 

For  example,  when  the  three  forces  P,  Q,  R  are  known, 
we  can  find  the  angles  which  the  cords  make  with  each 
other  when  in  equilibrium,  by  constructing  the  triangle 
ABC,  the  sides  of  which  are  proportional  to  these  forces. 
But  when  the  directions  are  given,  we  can  know  only 
the  ratios  of  the  three  forces ;  because  in  the  triangle 
ABC,  the  knowledge  of  the  three  angles  determines  only 
the  ratios  of  the  sides,  but  does  not  determine  their 
magnitudes.  Thus,  it  will  be  necessary,  besides,  to  know 
the  magnitude  of  one  of  the  three  sides  P,  Q,  R,  in  order 
to  find  that  of  the  two  others,  by  means  of  the  propor 
tional  series : 

P  :  Q  :  R  : :  AC  :  BC  :  AB. 

11 


122  STATICS. 


Remark  I. 

49-  128.  If  the  three  cords  be 

united  by  a  slip-knot ;  for  ex 
ample,  if  the  cord  PAQ  passes 
|       /          through  a  ring   attached   to 
/  the  extremity  of  the  cord  RA, 

^  the  conditions  just  enuncia 

ted  are  not  sufficient  to  es 
tablish  equilibrium  :  it  is  ne 
cessary,  in  addition,  that  the 
angles  PAB,  QAB,  formed  by 
the  two  parts  of  the  cord  and 
by  the  prolongation  AB  of  the  direction  of  the  other 
cord,  should  be  equal ;  for  it  is  evident  that,  otherwise, 
the  ring  A  would  slip  upon  this  cord  toward  the  greater 
of  the  two  angles. 


Remark  II. 

129.  What  has  just  been  said  contains  the  whole 
theory  of  equilibrium  betwee*n  three  powers  applied  to 
cords  united  in  the  same  knot ;  but  we  have  supposed 
the  construction  of  the  parallelogram  ACBD,  we  may 
however  enunciate  this  theory  independently  of  all  con 
struction. 

Thus,  in  every  triangle  ABC,  the  sides  are  proportional 
to  the  sines  of  the  opposite  angles :  that  is  to  say,  we 
have 

AC  :  BC  :  AB  : :  sin  ABC  :  sin  BAG  :  sin  ACB. 


CORDS.  123 

Now,  the  sines  of  these  angles  are  respectively  the 
same  as  those  of  their  supplements  RAQ,  RAP,  PAQ; 
hence  we  have  also 

AC  :  BC  :  AB  : :  sin  RAQ  :  sin  RAP  :  sin  PAQ, 
and  consequently 

p  :  Q  :  R  : :  sin  RAQ  :  sin  RAP  :  sin  PAQ. 

That  is  to  say,  when  the  three  powers,  which  act  by 
cords  upon  the  same  knot,  are  in  equilibrium,  each  of 
them  is  as  the  sine  of  the  angle  formed  ly  the,  directions 
of  the  other  two. 

COROLLARY  I. 

-  50-  130.  If  the  cords  AP,  AQ, 

instead  of  being  drawn  by 
two  powers,  be  attached  to 
two  fixed  points  at  P  and  Q, 
and  we  represent  the  force 
R  by  the  diagonal  AB  of  the 
parallelogram  ABCD,  the  two 
sides  AC,  AD  will  represent 
the  tensions  of  these  two  cords,  or  the  efforts  which 
they  exert  upon  the  fixed  points  in  the  line  of  their  di 
rections. 

COROLLARY  II. 

131.  When  the  angle  PAQ  is  very  large,  the  sides  AC, 
AD  of  the  parallelogram  are  very  great  compared  with 
the  diagonal  AB,  and  consequently  the  effort  which  the 


124 


STATICS. 


power  R  exerts  upon  the  fixed  points  P,  Q,  to  make  them 
approach  each  other,  is  very  great  compared  with  this 
power.  Hence,  by  means  of  cords,  we  can  put  a 
moderate  power  in  the  condition  of  producing  a  very 
great  effect. 

COROLLARY  III. 

132.  In  the  case  of  equilibrium,  however  small  the 
force  R  may  be,  the  diagonal  AB,  which  represents  it,  is 
not  zero,  and  the  three  points  c,  A,  D,  are  not  in  a  straight 
line ;  hence,  by  supposing  that  a  cord  PAQ  without  weight 
is  stretched  in  a  line  by  two  forces  P,  Q,  the  smallest 
force  R,  applied  to  A,  will  bend  it  at  this  point  and  cause 
it  to  make  an  angle  PAQ.  Thus,  it  is  rigorously  impossi 
ble  to  stretch  a  heavy  cord  in  a  straight  line,  unless  it 
be  vertical ;  for  the  weight  of  the  parts  which  compose 
it  may  be  regarded  as  forces  applied  to  this  cord,  and 
which  must  necessarily  deflect  it  from  a  straight  line. 


COROLLARY  IV. 


Fig.  51. 


133.  If  any  number 
of  powers  P,  Q,  R,  s, 
T,  .  .  .  act  upon  each 
other  by  cords  join 
ed  together  three  by 
three  in  the  same  knot, 
it  is  easy,  from  what 
precedes,  to  find  the 
ratios  which  these  pow- 


CORDS.  125 

ers  should  have  with  each  other  as  to  their  directions, 
so  as  to  be  in  equilibrium.  For  the  general  equilibrium 
cannot  take  place,  unless,  1st,  the  three  powers  joined 
together  in  each  knot  are  in  equilibrium  with  each  other ; 
2d,  each  of  the  cords  AB,  BC,  which  join  two  knots,  are 
equally  stretched  in  the  two  directions.  Hence,  by 
naming  v,  x,  the  tensions  of  the  two  cords  AB,  BC,  we 
will  have  (129),  by  reason  of  the  equilibrium  at  the 
knot  A, 

p  :  Q  : :  sin  QAB  :  sin  PAB, 
p  :  v  : :  sin  QAB  :  sin  PAQ  ; 

by  reason  of  the  equilibrium  at  the  knot  B, 

v  :  R  : :  sin  RBC  :  sin  ABC, 
v  :  x  : :  sin  RBC  :  sin  ABR  ; 

by  reason  of  the  equilibrium  at  the  knot  c, 


x  :  T  : :  sin  SCT  :  sm  BCS. 

And  by  continuing  these  proportions,  we  will  find  the 
ratio  of  any  two  of  these  powers,  and  the  ratio  of  one 
of  them  to  the  tension  v,  x  of  any  cord  whatever. 

For  example,  by  multiplying  in  order  the  2d  propor 
tion  and  the  3d,  we  find 

p  :  R  : :  sin  QABXsin  RBC  :  sin  PAQ  x  sin  ABC; 

by  multiplying  the  2d  and  4th, 
11* 


126  STATICS. 

p  :  x  : :  sin  QABXsin  RBC  :  sin  PAQXsin  ABR  ; 
by  multiplying  the  2d,  4th  and  5th, 

p  :  S  : :  sin  QABXsin  RBC X sin  SCT 
:  sin  PAQXsin  ABR  x  sin  BCT; 

by  multiplying  the  2d,  4th  and  6th, 

p  :  T  : :  sin  QABXsin  RBC  x  sin  SCT 
:  sin  PAQXsin  ABR  x  sin  BCS, 

and  so  on. 

It  also  follows  from  this,  that  the  three  cords  united 
by  the  same  knot  are  in  the  same  plane  (126),  although 
those  which  are  united  at  two  knots  may  be  in  different 
planes. 


COROLLARY  V. 


134.  If  the  forces  Q,  R,  s, 
be  weights  suspended  by  the 
knots  A,  B,  c,  to  the  same 
cord  EABCF,  and  this  cord 
be  retained  at  its  extremities 
by  two  fixed  points  E,  F  : 

1st.  The  main  co^d  and  the 
cords  of  the  weights  Q,  R,  s, 
are  in  the  same  vertical  plane ; 


CORDS.  127 

for  the  two  parts  EA,  AB  of  the  cord  are  in  the  vertical 
plane  drawn  through  the  cord  AQ  ;  in  like  manner,  the 
two  parts  AB,  BC  are  in  the  vertical  plane  drawn  through 
BR :  now  these  two  vertical  planes  pass  through  the 
same  line  AB,  and  coincide ;  hence  the  parts  EA,  AB,  BC 
of  the  cord,  and  the  directions  of  the  cords  AQ,  BR,  are 
in  the  same  vertical  plane.  In  the  same  manner  it  may 
be  demonstrated  that  the  part  CF  of  the  cord  and  the 
direction  CS  are  in  this  same  plane,  and  so  on. 

2d.  The  tensions  of  the  two  extreme  parts  of  the  cord 
are  to  each  other  reciprocally  as  the  sines  of  the  angles 
which  these  parts  make  with  the  vertical ;  for  the  angles 
QAB,  ABR,  are  supplements  of  each  other,  and  have  the 
same  sine :  it  is  the  same  with  the  angles  RBC,  BCS,  and 
so  of  the  rest ;  hence,  by  neglecting  the  common  factors 
in  the  proportion  which  gives  the  ratio  of  the  two  ex 
treme  tensions  P,  T  (133),  we  have 

E  :  T  : :  sin  SCT  :  sin  PAQ. 

3d.  The  vertical  HI,  drawn  through  the  point  of  inter 
section  G  of  the  prolongations  of  the  two  extreme  parts 
of  the  cord,  passes  through  the  centre  of  gravity  of  the 
system  of  all  the  weights  Q,  R,  s,  .  .  . ;  for  the  two  ex 
treme  parts  being  in  the  same  plane,  their  tensions  have 
a  resultant  whose  direction  passes  through  the  point  G ; 
moreover,  these  tensions  supporting  the  system  of 
weights  Q,  R,  S,  .  .  .  their  resultant  should  be  vertical, 
and  pass  through  the  centre  of  gravity  of  these  weights  ; 
hence  the  point  a  and  the  centre  of  gravity  of  the 
weights  Q,  R,  s,  are  in  the  same  vertical. 


128 


STATICS. 


COROLLARY  VI. 

135.  When  a  heavy  cord 
EHF  is  suspended  in  equilibri 
um  to  the  two  fixed  points 
E,  F,  we  may  consider  its  axis 
as  a  thread  without  weight, 
charged  with  a  weight  dis 
tributed  throughout  its  whole 
extent :  hence  1st,  this  axis  is 
in  the  vertical  plane  drawn  through  the  two  points  of 
suspension;  2d,  if  the  directions  of  the  two  extreme 
elements  of  this  axis  be  prolonged  to  EG,  FG,  and  through 
the  point  of  intersection  G  we  draw  the  vertical  IH,  the 
tensions  of  these  two  elements  are  to  each  other  recipro 
cally  as  the  sines  of  the  angles  which  these  elements 
make  with  the  vertical :  that  is  to  say,  by  naming  P  and 
T  these  tensions,  we  have 


p  :  T 


sin  IGF  :  sin  IGE 


3d,  the  centre  of  gravity  K  of  the  cord  is  in  the  ver 
tical  IH. 

Finally,  by  considering  the  total  weight  z  of  the  cord 
as  a  force  applied  to  the  point  G  of  its  direction,  we  will 
find  the  efforts  which  the  cord  makes  upon  the  two 
points  of  support  E,  F,  along  the  directions  EG,  FG,  by 
decomposing  the  force  z  into  two  others  which  act  in 
these  directions ;  and  we  shall  have  (129) 


z  :  P  :  T  : :  sm  EGF  :  sm  IGF  :  sin  IGE. 


COEDS.  129 


Remark. 

136.  So  far  we  have  supposed  that  there  were  only 
three  cords  united  at  each  knot,  because,  if  the  cords  as 
sembled  at  the  same  knot  are  greater  in  number  and 
included  in  the  same  plane,  it  is  not  sufficient  to  know 
their  directions  in  order  to  find  what  ratios  the  applied 
powers  should  have  so  as  to  be  in  equilibrium :  that  is 
to  say,  these  ratios  may  vary  in  an  infinite  number  of 
ways,  without  the  forces  ceasing  to  be  in  equilibrium. 

Thus,  whatever  may  be  the  number  of  powers  directed 
in  the  same  plane,  it  suffices,  in  order  to  be  in  equili 
brium  about  the  same  knot,  that  the  resultant  of  any 
two  of  them  is  equal  and  directly  opposed  to  the  re 
sultant  of  all  the  others ;  hence  all  these  forces,  except 
two,  being  taken  at  pleasure,  which  determine  the  in 
tensity  and  direction  of  the  resultant,  we  can  find  the 
intensities  of  the  latter  two  forces  which  will  make  an 
equilibrium  with  this  resultant. 

However,  when  four  cords,  joined  at  the  same  knot, 
are  not  in  the  same  plane,  their  directions  being  given, 
the  ratios  of  the  intensities  which  the  applied  forces 
must  have,  to  produce  an  equilibrium,  are  determined : 
for  we  have  seen  (44),  that  these  forces  must  be  to  each 
other  as  the  diagonal,  and  the  adjacent  sides  of  the 
parallelopipedon  constructed  on  their  lines  of  direction. 
But  when  the  forces  are  not  directed  in  the  same  plane, 
and  their  number  is  greater  than  four,  the  ratios  of  the 
forces  are  no  longer  determined  by  the  knowledge  of 
the  direction  of  the  cords. 


130 


STATICS. 


X 


ARTICLE  II. 

On  the  Equilibrium  of  the  Lever. 

Fig-  64.  137.  The  lever  is  an 

inflexible  rod  ACB  (Fig. 
54),  or  CAB  (Fig.  55), 
either  straight  or  curved 
and  moveable  around 
one  of  its  points  c,  ren 
dered  fixed  by  means  of 
any  obstacle ;  and  this 
obstacle  is  termed  the 
point  of  support  or  ful 
crum. 

138.  First,  supposing 
the  lever  to  be  without 
weight,  and  that  it  cannot 
in  any  manner  slip  upon 
its  fulcrum,  let  P,  Q  be  two 
powers  applied,  either  im 
mediately  or  by  means  of 
the  cords  AP,  BQ,  to  the 
two  points  A,  B  of  a  lever. 
If  we  consider  the  resist 
ance  of  the  point  c  as  the  effect  of  a  third  force  R  ap 
plied  to  the  lever  at  this  point,  we  have  seen,  in  order 
that  equilibrium  may  subsist  between  these  three  forces, 
1st,  their  directions  must  be  included  in  the  same  plane, 
and  meet  in  the  point  D  (10) ;  2d,  the  forces  P,  Q  must 
be  to  each  other  reciprocally  as  the  perpendiculars  CE, 


Fig.  65. 


THE   LEVER.  131 

CF,  let  fall  from  the  fulcrum  upon  their  directions  (35) : 
that  is  to  say,  we  must  have 

p  :  Q  : :  CF  :  CE  ; 

3d,  if  we  lay  off  from  the  point  D  the  lines  DL,  DM, 
upon  the  directions  of  the  forces  P,  Q,  proportional  to 
the  intensities  of  these  forces,  and  finish  the  parallelo 
gram  DLMN  ;  the  diagonal  DN  will  represent  in  intensity 
and  direction  the  force  E,  and  consequently  the  resist 
ance  of  the  fulcrum  (35) ;  thus  we  will  have 

p  :  Q  :  E  : :  DL  :  DM  or  NL  :  DN, 
or  (129), 

p :  Q  :  R  : :  sin  QBE,  :  sin  PDE  :  sin  PDQ. 


COEOLLAEY   I. 

139.  If  we  leave  out  of  consideration  the  resistance 
of  the  fulcrum :  that  is  to  say,  if  we  suppose  this  point 
to  be  capable  of  an  indefinite  resistance,  it  is  necessary, 
in  order  that  the  two  powers  P,  Q  may  be  in  equilibrium 
around  this  point  by  means  of  the  lever,  1st,  their  di 
rections  and  the  fulcrum  should  be  in  the  same  plane ; 
2d,  the  two  forces  P,  Q  should  tend  to  turn  the  lever 
around  the  fulcrum  c  in  opposite  directions,  and  their 
moments,  referred  to  this  point,  should  be  equal :  that 
is  to  say,  we  should  have  (80) 

PXCE=QXCF. 


132  STATICS. 


COROLLARY  II. 

140.  Hence  it  appears,  1st,  that  however  small  the 
power  Q  may  be,  we  may  always,  by  means  of  a  lever, 
put  it  in  equilibrium  around  a  fulcrum  c,  with  another 
power  P  of  given  intensity  and  direction ;  for  the  di 
rection  of  the  force  P  being  known,  the  distance  CE  of 
this  direction  from  the  fulcrum  will  be  known,  and  we 
will  know  the  moment  PXCE  :  hence  it  will  be  sufficient 
to  arrange  it  so  that  the  moment  QXCF  of  the  power,  is 
equal  to  the  preceding :  that  is  to  say,  to  direct  this 
power  in  such  a  manner  that  its  distance  CF  from  the 

P  X  CE 

fulcrum  is  equal  to ,  and  that  it  tends  to  turn  the 

lever  in  the  opposite  direction  to  the  force  P. 

2d.  If  the  distance  CF  of  the  direction  of  the  force 
Q  from  the  fulcrum  is  known,  we  will  find  the  intensity 
which  this  force  must  have  in  order  to  produce  an  equi 
librium  with  the  force  P,  by  dividing  the  moment  of  this 
latter  force  by  the  distance  CF  :  that  is  to  say,  we  shall 
have 

PXCE 


CF 


COROLLARY  III. 

141.  The  effort  or  load,  which  the  fulcrum  c  sustains, 
being  equal  to  the  resultant  of  the  two  forces  P,  Q,  we 
will  find  this  load  by  means  of  the  following  proportion  : 


THE   LEVER.  133 

p  :  Q  :  R  : :  sin  QDR  :  sin  PDR  :  sin  PDQ : 
that  is  to  say,  we  shall  have 

_PXsin  PDQ 
sin  QDR   ' 

or, 

_QXsin  PDQ 
sin  PDR 


COROLLARY  IV. 

142.  Hence,  if  the  fulcrum  does  not  possess  indefinite 
resistance,  in  order  that  it  may  not  be  moved,  and  that 
equilibrium  may  subsist,  it  is  necessary  that  the  resist 
ance  in  the  direction  CD,  should  be  equal  to  the  result 
ant  of  the  two  forces  P,  Q :  that  is  to  say,  equal  to 

PXsin  PDQ 


sn  QDR 


,  .  ,   .    ,,  QXsin  PDQ 

or,  which  is  the  same,  to  —  -.  —    -  . 

sm  PDR 


COROLLARY  V. 

143.  In  general,  of  the  six  things  to  be  considered 
in  the  equilibrium  of  the  lever,  namely  the  intensities 
and  directions  of  the  two  powers  p,  Q,  and  those  of  the 
load  on  the  fulcrum,  any  three  being  given,  the  other 

three  may  be  determined  in  all  cases;  or  of  the  six 

12 


134 


STATICS. 


analogous  things  which  may  be  considered  in  the  tri 
angle  DLN,  namely,  the  sides  and  the  angles,  three  being 
given,  we  can  determine  the  others. 

Remark  I. 

144.  If  the  lever  can  slip  upon  the  fulcrum,  the  con 
ditions,  which  have  just  been  given,  are  not  sufficient  to 
keep  the  lever  stable,  and  to  produce  an  equilibrium  ; 
it  is  necessary,  besides,  that  the  direction  DC  of  the  load 
on  the  fulcrum  should  be  perpendicular  to  the  surface 
of  the  lever  at  the  point  c  ;  for,  if  this  direction  were 
oblique,  the  lever  would  have  a  tendency  to  slip  towards 
the  side  of  the  greater  angle,  and  in  fact  would  slip, 
whenever  this  tendency  should  be  greater  than  the  fric 
tion  upon  the  fulcrum  which  opposes  this  effect,  as  we 
will  demonstrate  in  treating  of  the  inclined  plane. 


Remark  II. 

145.  What  has  just  been  stated  contains  the  whole 
theory  of  the  equilibrium  of  two  powers  applied  to  a 
lever,  considered  to  be  without  weight,  and  retained  by 
a  fulcrum ;  we  will  now  make  application  of  it  to  a  few 
simple  cases. 

Fiff-  56-  If  the  directions  of  the  pow 

ers  P,  Q,  applied  to  a  lever,  are 
parallel  to  each  other :  for  ex 
ample,  if  there  are  two  weights 
suspended  at  the  points  A,  B, 
the  load,  which  the  fulcrum  c 
sustains,  is  equal  to  their  sum 


THE   LEVER.  135 

P+Q,  and  the  two  perpendiculars  CE,  CF,  let  fall  from 
the  fulcrum  upon  their  directions,  are  in  a  straight  line. 
Hence,  if  the  lever  be  straight,  the  triangles  ACE,  BCF 
will  be  similar,  and  give 

CF  :  CE  : :  CB  :  CA. 

Hence  we  shall  have,  in  the  case  of  equilibrium, 
p  :  Q  : :  CB  :  CA  : 

that  is  to  say,  the  weights  P,  Q  will  be  to  each  other  re 
ciprocally  as  the  arms  of  the  lever. 

Thus,  the  intensity  and  the  lever  arm  of  a  resistance 
p  being  given,  1st,  the  lever  arm,  which  must  be  given 
to  a  power  Q  in  order  to  produce  an  equilibrium  with  it, 
will  be 


PXCA 
CB= — . 


2d.  The  intensity  of  the  power  necessary  to  be  ap 
plied  to  the  given  point  B,  in  order  to  produce  an  equi 
librium  with  it,  will  be 


_PXCA 

~~* 


Finally,  if  the  two  weights  P,  Q,  and  the  length  AB 
of  the  lever,  be  given,  we  will  find  the  fulcrum  c,  around 
which  these  weights  will  be  in  equilibrium,  by  dividing 
the  lever  AB  into  two  parts  reciprocally  proportional  to 
the  two  weights. 


136  STATICS. 


Remark  III. 

146.  When  there  are  more  than  two  powers  applied 
to  the  same  lever,  it  is  not  sufficient  to  know  their  di 
rections,  and  the  position  of  the  fulcrum,  in  order  to 
determine  the  ratios  which  they  should  have  to  each 
other  so  as  to  be  in  equilibrium ;  but,  as  equilibrium 
cannot  take  place  between  several  forces  about  a  ful 
crum,  unless  the  resultants  of  all  the  forces  are  destroyed 
by  the  resistance  of   this  point,  it  is  evident,  in  this 
case,  that  the  conditions  of  equilibrium  are  reduced  to 
the  two  following  :  1st,  all  the  forces  must  have  a  single 
resultant ;  2d,  the  direction  of  this  resultant  must  pass 
through  the  fulcrum. 

COROLLARY. 

147.  If  the  directions  of  all  the  forces  are  included 
in  the  same  plane,  these  forces  have  necessarily  a  single 
resultant  (43),  and  the  first  condition  is  fulfilled ;  hence 
it  suffices  for  the  equilibrium,  that  the  direction  of  this 
resultant  passes  through  the  fulcrum,  or,  which  amounts 
to  the  same,  the  sum  of  the  moments  of  the  forces, 
which  tend  to  turn  the  lever  in  one  direction  around  the 
fulcrum,  is  equal  to  the  sum  of  the  moments  of  those 
which  tend  to  turn  it  in  the  opposite  direction. 

Remark  IV. 

148.  So   far  we   have   abstracted  weight  from  the 
lever ;  but  if  this  weight  enter  into  consideration,  it 
must  be  regarded  as  a  new  force,  applied  to  the  centre 


THE   LEVER.  137 

of  gravity  of  the  lever  in  a  vertical  direction ;  and  in 
the  case  of  equilibrium,  the  conditions  just  stated  sub 
sist  between  all  the  forces,  including  that  under  con 
sideration. 

Let  P,  Q  be  two  weights  sus 
pended  from  a  heavy  lever  AB, 
and  in  equilibrium  about  the 
fulcrum  c  :  we  will  consider  the* 
weight  of  the  lever  as  a  third 
weight  S,  suspended  from  the 
centre  of  gravity  of  the  lever, 
and  the  sum  of  the  moments  of  the  two  weights  Q,  S, 
referred  to  the  fulcrum  c,  will  be  equal  to  the  moment 
of  the  weight  P :  that  is  to  say,  we  shall  have 

QXCB+SXCG=PXCA; 

or,  subtracting  from  these  two  equal  quantities  the  mo 
ment  of  the  lever  SXCG, 

QXCB=PXCA— SXCG. 

Thus,  knowing  the  length  and  the  weight  of  the  lever, 
the  position  of  its  centre  of  gravity,  that  of  the  fulcrum 
and  one  of  the  two  weights  P,  Q,  it  will  always  be  pos 
sible  to  get  the  other  weight ;  for  we  shall  have 

_QXCB-fSXCG 
CA  9 

and 

PXCA— SXCG 


Q= 


CB 

12* 


138 


STATICS. 


The  load  at  the  fulcrum  is  evidently  equal  to  the  sum 
of  the  weights  p-fQ+s. 

F^-  58<  149.  But,  if  the  weight  P, 

suspended  from  the  heavy 
lever  AB,  be  retained  in  equi 
librium  about  the  fulcrum  c,  by 
means  of  a  vertical  power  Q, 
and  directed  upwards,  the  mo 
ment  of  the  force  Q,  which 
tends  to  turn  the  lever  in  one 
direction,  will  be  equal  to  the 
sum  of  the  moments  of  the 
weights  P,  s,  which  tend  to 
turn  it  in  the  opposite  direc 
tion,  and  we  shall  have 

QXCB=PXCA  +  SXCG, 

or,  subtracting  the  moment  of  the  lever, 

QXCB  — SXCQ=PXCA. 

Thus  the  intensities,  which  the  two  forces  P,  Q  must 
have  in  order  to  be  in  equilibrium,  will  be 


_ 


QXCB—  SXCG 
~CA 


and 


_ 


PXCA+sxca 

CB         ' 


and  the  weight  upon  the  fulcrum  p+S—  Q. 


THE   LEVER. 


139 


O  tlOLLARY. 

Fig.  57.  150.    Hence  it  appears  by 

regarding  the  weight  P  as  a  re 
sistance,  and  the  force  Q  as  a 
power  which  has  to  bring  the 
weight  into  equilibrium  or  to 
overcome  it,  by  means  of  the 
lever  AB,  the  weight  of  this 
lever  is  a  force  which  may  either  increase  or  diminish 
the  power,  according  as  this  weight  tends  to  turn  the 
lever  in  the  same  direction  as  the  power,  or  in  the  oppo 
site  direction.  For  example,  in  the  case  of  Fig.  57, 
the  weight  of  the  lever  increases  the  power  Q ;  and  by 
prolonging  the  lever  arm  CB  of  this  power,  it  would  be 
placed  in  the  condition  of  producing  an  equilibrium  with 
a  greater  resistance,  for  two  reasons :  1st,  because  its 
Fig-  58.  moments  would  be  thus  aug- 

^  mented;  2d,  because  the 
weight  S  of  the  lever  would 
be  augmented,  which  alone 
would  produce  an  equilibrium 
with  a  greater  part  of  the 
resistance.  But,  in  the  case 
of  Fig.  58,  the  weight  of 
the  lever  diminishes  the  pow 
er  Q,  and  we  cannot  increase 
the  length  of  the  lever  arm 
*"  CB  without  at  the  same  time 

augmenting  its  weight  S,  which  forms  part  of  the  resist 
ance  ;  thus,  in  order  that  there  may  be  in  this  case  an 


I 

A 
v 


*''"*   ^     j* 

"' 


140  STATICS. 

advantage  in  prolonging  the  arm  of  the  lever,  it  is 
necessary  that  the  moment  of  this  prolongation  should 
be  less  than  the  resulting  increase  in  the  moment  of  the 
power,. 


THEOREM. 

151.  Two  powers  p, 
Q,  api^lied  to  a  lever 
AB,  and  in  equilibrium 
about  a  fulcrum  c,  are 
t°  each  other  recipro- 
cally  as  tlie  spaces 
which  these  powers  tra- 
verse  in  the  line  of 
their  directions  ,  if  the 
equilibrium  be  disturbed  infinitesimally  . 

DEMONSTRATION.  From  the  fulcrum  c  let  fall  the 
perpendiculars  CE,  CF  upon  the  directions  of  the  powers  ; 
and  instead  of  the  straight  lever  AB,  let  us  consider  the 
bent  lever  ECF  ;  at  the  extremities  E,  F  of  which  con 
ceive  the  powers  P,  Q  to  be  applied  ;  then  suppose,  by 
virtue  of  a  derangement  in  the  equilibrium,  that  the 
bent  lever  EOF  takes  the  infinitely  near  position  ecf. 
This  being  done,  the  small  arcs  EC,  F/,  will  be  the  spaces 
which  the  powers  P,  Q  would  traverse  by  virtue  of  this 
derangement.  Now,  the  angle  ECF  of  the  bent  lever 
being  invariable,  the  two  angles  EC<?,  FC/  are  equal,  and 
we  have 


THE   LEVER. 
CF  :  CE  :  :  F/  :  Ee  ; 

moreover,  because  of  the  equilibrium,  we  have  (35) 

p  :  Q  :  :  CF  :  CE  ; 
hence  we  have  also 


141 


We  will  have  occasion  to  show,  in  a  subsequent  part, 
that  an  analogous  proposition  occurs  in  cases  of  equi 
librium  for  all  other  machines. 


On  Pulleys. 


Fig.  60. 


I. 


152.  A  pulley  is  a 
wheel  having  a  groove  on 
its  circumference  to  re 
ceive  a  cord  PGDHQ,  and 
is  traversed  at  its  centre 
by  an  axle  E,  upon  which 
it  turns  in  a  sheath  or 
block  EL. 


142 


STATICS. 


II. 

Fig.  60.  153.  Suppose  the  axis 

of  the  pulley  being  fixed, 
two  forces  p,  Q  are  ap 
plied  to  the  extremities 
of  the  cord,  and  this  cord 
being  perfectly  flexible 
exerts  no  friction  upon 
the  groove  of  the  pulley, 
so  that  it  may  slide  free 
ly  on  this  rim.  What 
ever  the  figure  of  the 
pulley  may  be  in  other 
respects,  that  is  to  say, 
whether  the  arc  GDH,  em 
braced  by  the  cord,  be  circular  or  not,  it  is  evident  that 
the  two  forces  cannot  be  reciprocally  in  equilibrium 
unless  they  are  equal ;  for  if  they  were  unequal,  the 
greater  would  overcome  the  smaller  by  causing  it  to 
slide  in  the  groove  of  the  pulley. 

Upon  the  same  supposition,  the  pulley,  having  no 
other  fixed  point  than  its  centre,  and  being  drawn  by 
the  two  forces  P,  Q,  cannot  remain  at  rest,  unless  the 
resultant  of  these  two  forces  is  directed  towards  the 
centre,  and  is  destroyed  by  the  resistance  of  this  point. 
Hence,  having  prolonged  the  directions  PG,  QH  of  the 
two  cords,  until  they  meet  in  some  point  A,  and  taking 
upon  these  directions  the  equal  lines  AB,  AC  to  repre 
sent  the  forces  P,  Q,  if  the  parallelogram  ABDC  be  com 
pleted,  the  diagonal  AD,  .which  will  represent  the  resultant 
of  these  two  forces,  must  pass  through  the  fixed  point  E. 


PULLEYS. 


143 


Fig.  61. 


Now,  when  the  figure 
is  circular,  this  last  con 
dition  is  always  fulfilled : 
thus,  the  triangle  ABD 
being  isosceles,  the  angle 
BAD  is  equal  to  the  angle 
BDA,  and  consequently  to 
the  angle  DAG  ;  hence  the 
diagonal  AD  divides  the 
angle  BAG  into  two  equal 
parts.  But,  if  we  draw 
the  line  EA,  this  line  di 
vides  the  same  angle  into 
two  equal  parts ;  for  if, 
through  the  centre  E,  we  draw,  to  the  points  of  contact 
of  the  cords,  the  radii  EG,  EH,  perpendicular  to  the  di 
rections  of  these  cords,  the  two  rectangular  triangles 
EGA,  EHA  will  be  in  all  respects  equal,  and  we  shall  have 
the  angle  EAG  of  the  one,  equal  to  the  angle  EAH  of  the 
other.  Hence  the  line  EA,  and  the  diagonal  DA,  will 
have  the  same  direction. 

Hence,  the  centre  of  a  pulley  being  fixed  and  its 
figure  circular,  it  is  necessary  that  the  two  forces  P,  Q, 
to  be  in  equilibrium,  should  be  equal,  and  that  the  pulley 
remain  at  rest,  at  the  same  time,  about  its  axis. 

The  load  which  the  axis  of  the  pulley  sustains  is  evi 
dently  equal  to  the  resultant  of  the  two  forces  P,  Q; 
hence,  if  we  name  R  this  load,  we  shall  have  (36) 


p  :  Q  :  R  : :  AB  :  AC  :  AD. 


144 


STATICS. 


Fig.  60. 


Finally,  draw  the  cord 
GH  of  the  arc  embraced 
by  the  rope  ;  the  two  tri 
angles  GHE,  ABD  will  be 
similar,  because  they  will 
have  their  sides  perpen 
dicular  each  to  each ;  and 
we  shall  have 

AB  :  AC  :  AD  : :  GE  :  EH  :  GH; 

hence  we  shall  have 

p  :  Q  :  n  : :  GE  :  EH  :  GH. 


III. 


154.  If  the  axis  of  the  pulley  is  not  absolutely  fixed, 
but  retained  simply  by  the  power  s,  by  means  of  the 
block  EL  and  the  cord  LS  ;  in  order  that  this  axis  may 
be  at  rest,  and  the  three  forces  P,  Q,  S  in  equilibrium, 
it  is  necessary  that  the  force  S  should  be  equal  and  di 
rectly  opposed  to  the  load  which  the  axis  sustains. 
Hence,  1st,  the  direction  of  this  force  should  coincide 
with  the  line  EA  ;  2d,  its  intensity  should  be  equal  to 
the  resultant  n  of  the  two  forces  P,  Q,  and  we  shall 
have 


p  :  Q  :  S  : :  GE  :  EH  :  GH. 


PULLEYS. 


145 


Thus,  when  two  forces  P,  Q,  applied  to  a  rope  em 
bracing  a  pulley,  are  in  equilibrium  with  each  other,  and 
with  a  third  force  S  applied  to  the  axis  of  the  pulley, 
1st,  the  two  forces  P,  Q  are  equal  to  each  other  ;  2d,  the 
direction  of  the  force  S  bisects  the  angle  formed  by  the 
directions  of  the  two  others ;  3d,  each  of  the  two  forces 
P  and  Q  is  to  the  third  force  S,  as  the  radius  of  the  pul 
ley  is  to  the  chord  of  the  arc  embraced  by  the  rope. 


COROLLARY  I. 


Fig.  60.  155.  We  observe  that 

if  the  cord,  fastened  to 
the  block,  instead  of  being 
drawn  by  a  force  S,  is  at 
tached  to  a  fixed  point  of 
indefinite  resistance,  and 
if  it  be  proposed,  by  em 
ploying  the  pulley,  only 
to  place  the  two  forces 
P,  Q  in  equilibrium,  or  to 
overcome  a  resistance  P, 
by  means  of  a  power  Q, 
the  pulley  does  not  assist 
the  power :  it  has  no 

other  effect  than  to  change  the  direction  of  this  force, 

without  altering  its  intensity. 


is 


146 


STATICS. 


ir.  61. 


But  if  one  of  the  ex 
tremities  of  the  cord, 
which  embraces  the  pul 
ley,  is  attached  to  a  fixed 
point  P,  and,  by  using 
the  pulley,  we  design  to 
place  a  power  Q  in  equi 
librium  with  a  resistance 
S,  attached  to  the  block, 
the  pulley  assists  the 
power,  which  is  always 
less  than  the  resistance ; 
for  we  have 


Q  :  S  : :  EH  :  GH, 


COROLLARY  II. 


Fig,  63. 


156.  When  the  directions 
of  two  parts  PG,  QH  of  the  cord 
are  parallel  to  each  other,  and 
consequently  to  that  of  the 
cord  of  the  block,  the  chord 
GH  becomes  a  diameter,  and 
is  double  the  radius  ;  the  pro 
portional  series  of  No.  153 
then  becomes 

p  :  Q  :  S  : :  1  :  1  :  2, 


that  is  to  say,  the  force  S,  or  the  load  of  the  axis  of  the 
pulley,  is  equal  to  the  sum  of  the  two  powers  P,  Q,  or  to 


PULLEYS.  147 

double  one  of  them.  Thus,  in  the  case  of  Fig.  63,  the 
power  Q,  which,  by  means  of  the  pulley  and  the  point 
of  support  P,  will  produce  an  equilibrium  with  the  re 
sistance  S,  will  be  only  half  of  this  resistance. 


IV. 

157.  A  pulley  is  said  to  be  immoveable  when  its 
block  is  attached  to  a  fixed  point  and  the  power  and  re 
sistance  are  applied  to  the  cord  which  embraces  it ;  but 
when  the  resistance  is  attached  to  the  block,  and  the 
pulley  has  to  move  with  it,  as  in  Figs.  61,  63,  the  pulley 
is  said"  to  be  movedble. 

This  being  established,  let  there  be  any  number 
of  moveable  pulleys,  (Fig.  64,)  and  considered  without 
weight ;  let  the  first  bear  a  weight  P  suspended  to  its 
hook,  and  embraced  by  a  cord,  one  of  whose  extremi 
ties  is  attached  to  the  fixed  point  D,  and  the  other  is 
applied  to  the  block  of  the  second  pulley ;  let  this  pul 
ley  be  embraced  by  another  cord,  one  of  whose  extremi 
ties  is  fixed  to  the  point  H,  and  the  other  attached  to 
the  block  of  the  next  pulley ;  let  this  third  pulley  be 
embraced  by  a  third  cord  fixed  on  one  side  to  M,  and 
drawn  on  the  other  by  a  power  Q  ;  and  so  on,  if  the 
number  of  pulleys  be  greater.  Finally,  supposing  the 
whole  system  to  be  in  equilibrium,  draw  the  radii  and 
the  subtending  chords  of  the  pulleys  as  is  shown  in  the 
figure.  We  may  consider  the  equilibrium  of  the  first 
pulley  A  as  though  this  pulley  were  alone ;  and  represent 
ing  by  X  the  tension  of  the  cord  BX,  we  shall  have  (155) 

p  :  x  : :  BC  :  BA. 


148 


STATICS. 
Fig.  64. 


For  the  same  reason,  calling  Y  the  tension  of  the  cord 
FT,  -we  shall  have 

x  :  Y  : :  FG  :  EF. 

We  shall  have,  in  like  manner,  for  the  third  pulley, 
Y  :  Q  : :  KL  :  IK  ; 

and  so  on,  whatever  may  be  the  number  of  pulleys. 
Hence,  by  multiplying  in  order  all  these  proportions, 
we  shall  have 

p  :  Q  : :  BCXFQXKL  :  BAXEFXIK: 


PULLEYS. 


149 


that  is  to  say,  the  resistance  is  to  the  power  as  the 
product  of  the  subtending  chords  is  to  the  product  of 
the  radii. 


COROLLARY. 

158.  When  all  the  ropes  CD,  GH,  LM,  .  .  are  parallel, 
the  subtending  chords  will  be  diameters,  and  the  pre 
ceding  proportion  will  become 

p  :  Q  ::  2x2x2  :  Ixlxl,  or  : :  8  :  1 ; 


Fig.  65. 


from  which  it  is  evident,  that, 
in  this  case,  the  resistance  is  to 
the  power  as  the  number  2, 
raised  to  a  power  indicated  by 
the  number  of  moveable  pul 
leys,  is  to  unity. 

Thus,  by  suitably  increasing 
the  number  of  moveable  pulleys, 
we  may  put  a  moderate  force  in 
equilibrium  with  a  very  great 
resistance.  For  example,  with 
three  pulleys,  and  by  means  of 
the  fixed  points  D,  II,  M,  the 
power  produces  an  equilibrium 
with  a  resistance  eight  times 
greater  than  itself. 

However  advantageous  this 
disposition  of  moveable  pulleys 
may  at  first  appear,  it  is  rarely 


13* 


150  STATICS. 

employed ;  because,  in  order  to  make  the  first  pulley  A 
traverse  a  certain  space,  it  is  necessary  that  the  second 
should  traverse  a  double  space,  and  the  third  a  quadru 
ple  ;  and  so  on,  which  requires  too  much  room ;  and 
thus,  most  generally,  muffles  are  used  instead. 


V. 


159.  The  term  muffle  is  applied  to  a  system  of  several 
pulleys  assembled  in  the  same  block  and  upon  separate 
axes,  as  in  Figs.  66,  67,  or  upon  the  same  axis,  as  in 
Fig.  68.  A  fixed  muffle  and  a  moveable  muffle  are 
always  employed  at  the  same  time ;  and  all  the  pulleys 
of  the  two  muffles  are  embraced  by  the  same  cord,  one 
of  whose  extremities  is  attached  to  one  of  the  two 
muffles,  and  the  other  extremity  is  drawn  by  the  power ; 
and  the  resistance  is  suspended  to  the  moveable  muffle. 

We  may  give  different  diameters  to  the  pulleys,  and 
arrange  them  in  such  a  manner  that  all  the  parts  of  the 
cord,  which  go  from  one  muffle  to  the  other,  may  be 
parallel  to  each  other,  as  in  Figs.  66,  67 ;  but  this  ar 
rangement  increases  the  extent  of  the  muffles.  They 
may  be  reduced  to  a  volume  much  smaller  and  more 
convenient,  by  mounting  in  each  of  them  all  the  pulleys 
upon  the  same  axis,  as  in  Fig.  68.  By  this  means,  the 
cords  which  are  on  one  side  of  the  muffles  are  not 
parallel  to  those  on  the  other  side ;  but,  when  the 
distance  of  the  muffles  is  considerable,  the  departure 
from  parallelism  is  very  small,  and  it  may  be  regarded 
as  insensible. 


PULLEYS. 


151 


Fig. 


Fig.  68. 


160.  By  considering,  then,  the  cords  of  the  muffles 
as  parallel  to  each  other,  and  abstracting  all  weight 
from  the  whole  machine,  let  Q  be  a  power  in  equilibrium 
with  the  resistance  P,  suspended  from  the  block  of  the 
moveable  muffle.  Equilibrium  cannot  exist  throughout 
the  whole  machine,  unless  it  occur  in  each  particular 
pulley,  and  the  two  parts  of  the  cord  which  embraces 
this  pulley  are  equally  stretched  (154).  Thus,  the  ten 
sions  of  the  two  cords  QA,  BC  are  equal  to  each  other ; 
so  likewise  are  those  of  the  two  cords  BC,  DE,  and  those 


152 


STATICS. 


A 


Fig.  6G.  Of  the  cords  DE,  FG,  and  so  on,  what- 

||   ever  may  be  the  number  of  the  cords ; 


hence  all  the  cords  which  go  from  one 
5  muffle  to  the  other  are  equally  stretch 
ed.  Now  the  sum  of  these  tensions 
produces  an  equilibrium  with  the  re 
sistance  P,  and  is  equal  to  it ;  or,  what 
is  the  same,  the  tension  of  one  of  the 
cords,  multiplied  by  their  number,  is 
equal  to  the  resistance  ;  hence  the  ten 
sion  of  one  of  the  cords,  or  the  power 
Q,  is  the  quotient  of  the  resistance  p  di 
vided  by  the  number  of  cords  which  go 
from  one  muffle  to  the  other. 

Hence  it  is  evident,  that,  in  the  case 
of  Fig.  66,  where  the  extremity  of  the 
cord  is  attached  to  the  fixed  muffle,  the 
power  Q  should  be  one-sixth  of  the  re 
sistance  to  be  in  equilibrium  with  it,  and 
in  the  case  of  Fig.  67,  where  the  ex 
tremity  of  the  cord  is  attached  to  the 
moveable  muffle,  it  should  be  one-fifth, 

because  there  is  one  pulley  and  one  cord  less. 

If  it  be  desired  to  introduce  the  consideration  of  the 

weight  of  the  moveable  muffle,  it  may  be  regarded  as 

constituting  part  of  the  resistance. 


WHEEL   AND   AXLE. 


153 


Of  the  Wheel  and  Axle. 

I. 

161.  The  wheel  and  axle,  windlass,  or  capstan,  is  a 
machine  consisting  of  a  cylinder  moveable  upon  an  axis, 
and  of  a  cord,  which,  having  one  of  its  extremities 
wound  around  the  cylinder,  while  a  power  Q  causes  it 
to  turn,  overcomes  a  resistance  P  attached  to  its  other 
extremity.  The  cylinder  is  furnished  at  its  two  ends 
with  trunnions  A,  B,  bearing  upon  supports,  which  are 
named  boxes,  and  by  means  of  which  it  turns  freely  on 
its  axis. 

Fig.  69. 


162.  There  are  several  modes  of  applying  the  power 
to  this  machine,  to  communicate  the  movement  of  rota 
tion  to  the  cylinder. 


154  STATICS. 

1st.  We  may  unite  solidly  with  the  cylinder  and  upon 
the  same  axis,  a  wheel,  whose  circumference,  having  a 
groove  in  it  like  that  of  a  pulley,  is  surrounded  by  a 
second  cord.  This  cord,  being  drawn  by  the  power, 
causes  both  the  wheel  and  the  cylinder  to  turn  upon 
their  common  axis.  This  first  arrangement,  to  which  we 
will  refer  all  the  others,  is  rarely  employed,  because  it 
requires  the  cord  of  the  wheel  to  be  very  long,  while 
the  space  to  be  traversed  by  the  resistance  is  incon 
siderable. 

2d.  The  rim  of  a  wheel,  furnished  with  spokes  placed 
equally  far  apart,  and  to  which  men's  hands  are  applied, 
furnishes  them  with  the  means  of  making  use  of  part 
of  their  weight  in  turning  the  machine. 

3d.  In  other  cases,  instead  of  the  wheel,  there  is 
mounted  upon  the  cylinder  a  large  hollow  drum  in  which 
men  or  animals  can  walk ;  and  then  by  their  feet  they 
cause  both  the  drum  and  the  cylinder  to  turn. 

4th.  Sometimes,  in  place  of  making  use  of  the  wheel 
and  the  drum,  the  cylinder  is  traversed  by  bars  perpen 
dicular  to  its  axis,  and  at  the  extremities  of  which  men 
act  by  the  force  of  their  muscles  and  part  of  their 
weight. 

5th.  Finally,  most  frequently,  and  when  the  resist 
ance  is  not  very  great,  there  are  adapted  to  the  ex 
tremities  of  the  cylinder  one  or  two  cranks,  which  men 
turn  by  employing  the  force  of  their  arms. 

163.  The  names  of  this  machine  vary  with  its  object, 
and  even  with  its  position.  Ordinarily  it  is  named  wind 
lass,  and  wheel  and  axle,  when  the  cylinder  is  horizontal ; 
and  capstan,  when  the  cylinder  is  vertical,  and  horizon 
tal  bars  are  used  to  put  it  in  motion. 


WHEEL   AND   AXLE. 


155 


It  is  evident  that  the  different  modes  in  which  the 
power  may  be  applied  to  the  cylinder  of  the  wheel  and 
axle,  can  all  be  referred  in  theory  to  the  first  we  have 
described;  for,  whatever  may  be  the  direction  of  the 
power,  when  it  is  directed  in  a  plane  perpendicular  to 
the  axis  of  the  cylinder,  we  may  always  conceive  it  to 
be  applied  to  a  wheel  whose  circumference  would  be 
tangent  to  the  direction  of  this  power.  Thus  we  will 
suppose  that  VXYZ  being  the  cylinder  of  the  machine, 
whose  axis  ACB  is  perpendicular  to  the  plane  of  the 
wheel  JiDD'Jc,  1st,  the  power  Q  is  applied  to  the  circum 
ference  of  this  wheel  in  any  direction  DQ,  contained  in 
the  plane  of  the  wheel,  and  tangent  to  the  circumference 
of  the  radius  CD,  at  a  given  point  D ;  2d,  that  the  di 
rection  KP  of  the  resistance  is  tangent  at  K  to  the  sur 
face  of  the  cylinder,  and  situated  in  a  plane  parallel  to 
that  of  the  wheel.  Lastly,  to  render  the  conception 
clear,  we  will  suppose  that  the  axis  AB  of  the  cylinder 
is  horizontal,  and  consequently  that  the  direction  KP  of 


156  STATICS. 

the  resistance  is  vertical.  This  being  granted,  two 
questions  present  themselves  to  be  solved :  the  first  is 
to  find  the  ratio  which  the  power  Q  and  the  resistance  P 
should  have  to  produce  an  equilibrium  ;  the  second  is  to 
find  the  weight  which  the  supports  of  the  two  pivots 
A,  B,  sustain. 


II. 

164.  To  solve  the  first  of  these  two  questions,  let  us 
conceive  a  horizontal  plane  KMEN  to  pass  through  the 
axis  AICB  of  the  cylinder :  this  plane  will  pass  through 
the  point  K,  where  the  direction  of  the  resistance  touches 
the  surface  of  the  cylinder ;  moreover,  it  will  intersect 
the  plane  of  the  wheel  in  a  horizontal  line  ME,  which 
will  pass  through  the  centre  c,  and  it  will  meet  the  di 
rection  of  the  power  Q  in  some  point  E.  Prolong  the 
line  ME  to  ES,  and  through  the  point  E  draw  the  vertical 
ER;  the  three  lines  EQ,  ER,  ES  being  included  in  the 
same  plane,  which  is  that  of  the  wheel,  we  can  decom 
pose  the  power  Q  into  two  forces  R,  S,  directed  along  EG, 
EH.  For  this  purpose,  we  will  represent  this  power  by 
the  parts  EF  of  its  direction  ;  and  completing  the  paral 
lelogram  EGFH,  we  will  have 

Q  :  S  : :  EF  :  EH, 

Q  :  R  : :  EF  :  EG,  or  FH  ; 

or,  since,  by  drawing  the  radius  CD,  the  two  rectangular 
triangles  CDE,  EHF  will  be  similar,  and  give 


EF  :  FH  :  EH  : :  CE  :  CD  :  DE 


we  shall  have 


WHEEL   AND   AXLE. 


Q  :  S  : :  CE  :  DE, 
Q  :  E  : :  CE  :  CD. 


15T 


Thus,  in  place  of  the  power  Q,  we  can  take  the  two 
forces  E,  S,  whose  values 


QXCD 
E=- , 

CE    ' 


QXDE 


are  known,  since  the  direction  of  the  force  Q  being 
given,  all  parts  of  the  triangle  CDE  are  known. 
Fig.  69. 


Now,  of  the  two  forces  E,  S,  the  latter  being  directed 
towards  the  axis  of  the  cylinder  which  is  immoveable, 
it  may  be  regarded  as  immediately  applied  to  the  point 


158  STATICS. 

c,  and  destroyed  by  the  resistance  of  the  axis ;  this 
force,  then,  cannot  contribute  in  any  manner  to  the 
motion  of  rotation  of  the  cylinder,  and  it  has  no  other 
effect  than  that  of  pressing  the  pivots  against  their 
supports.  Hence,  there  remains  only  the  force  R  which 
can  be  employed  to  produce  an  equilibrium  with  the  re 
sistance  P. 

Now,  in  the  horizontal  plane  KMEN  draw  the  radius 
of  the  cylinder  KI,  which  will  be  perpendicular  to  the 
axis  and  parallel  to  ME  ;  also  draw  the  line  KE  which 
will  intersect  the  axis  in  some  point  L.  This  being 
done,  the  point  L,  being  in  the  axis,  may  be  regarded  as 
immoveable,  and  the  line  KE  may  be  taken  for  an  inflexi 
ble  rod,  retained  by  a  fulcrum  L,  at  the  extremities  of 
which  are  applied  the  two  forces  R,  P.  Now  the  direc 
tions  of  these  two  forces  are  both  vertical,  and  conse 
quently  parallel ;  hence  in  order  that  equilibrium  may 
subsist  between  them,  it  is  necessary  that  they  should 
be  reciprocally  proportional  to  their  arms  LE,  LK,  or 
that  we  should  have 

R  :  p  : :  KL  :  LE. 
But  the  similar  rectangular  triangles  KIL,  LCE,  give 

KL  :  LE  : :  KI  :  CE  ; 

hence  we  shall  have 

R  :  p  : :  KI  :  CE. 


WHEEL   AND   AXLE.  159 

Hence  by  multiplying  in  order,  this  proportion  and 
the  following 

Q  :  R  : :  CE  :  CD, 

which  we  found  above,  we  shall  have,  in  case  of  equi 
librium, 

* 

Q  :  p  : :  KI  :  CD  ; 

that  is  to  say,  the  power  will  be  to  the  resistance  as  the 
radius  of  the  cylinder  is  to  the  radius  of  the  wheel; 
which  is  the  answer  to  the  first  question. 

By  making  the  product  of  the  extremes  equal  to  that 
of  the  means  in  the  above  proportion,  we  have 

QXCD  =  PXKI. 

The  line  CE  meets  the  circumference  of  the  wheel  at 
the  point  D';  imagining  that  there  is  through  this  point 
a  force  Q'  equal  to  Q  directed  along  the  tangent  D'Q',  we 
shall  have  (31) 

QXCD=Q'XCD/=PXKI  ; 

from  which  it  appears,  that  in  the  case  of  equilibrium, 
the  moments  of  the  power  Q',  equal  to  Q,  and  of  the  re 
sistance  P,  both  taken  with  reference  to  the  vertical 
plane  passing  through  the  axis  of  the  cylinder,  are  equal 
to  each  other. 


160 


STATICS. 


III. 

165.  The  pressure  which  the  trunnions  exert  against 
their  points  of  support  is  evidently  the  effect  only  of 
the  forces  P,  Q,  and  of  the  weight  T  of  the  machine, 
which  may  be  considered  as  united  at  its  centre  of 
gravity  g,  where,  by  taking  the  two-  forces  R,  S,  instead 
of  the  power  Q,  these  pressures  are  produced  by  the  four 
forces  P,  E,  s,  T. 

Fig.  69. 


These  forces  are  all  known ;  for,  1st,  the  resistance 
p  and  the  weight  T  of  the  machine  are  given  imme 
diately  ;  2d,  the  two  other  forces  R  and  s,  whose  values 
we  have  found  to  be,  in  general, 


QXCD 


B=«*5?, 

CD     ' 


WHEEL   AND   AXLE.  161 

become,  in   the    case   of  equilibrium   where   we   have 
QXCD=PXKI, 


s= 


PXKI 
R= , 

CE 

PXKIXDE 


CDXCE     : 


and  containing  only  known  quantities,  since  the  direc 
tion  of  the  power  Q  being  given,  we  can  find  all  the 
parts  of  the  right-angled  triangle  CDE. 

Now  the  two  forces  P,  R,  whose  directions  are  vertical, 
and  which  are  in  equilibrium  about  the  point  L,  exert 
upon  this  point  of  the  axis  a  weight  whose  direction  is 
vertical,  and  which  is  equal  to  their  sum  P+R.  More 
over,  this  weight  P+R,  being  sustained  by  the  two  points 
of  support  at  A  and  B,  must  be  regarded  as  the  resultant 
of  the  two  vertical  pressures  which  it  exerts  at  these 
points ;  and  we  will  find  each  of  these  pressures,  by 
dividing  the  resultant  P+R  into  two  parts  reciprocally 
proportional  to  the  distances  of  the  point  L  from  the 
two  supports.  Let  x  be  the  pressure  exerted  at  the 
point  A,  and  x'  that  exerted  at  the  point  B ;  we  will  find 
these  two  pressures  by  the  following  proportions : 

AB  :  BL  : :  P+R  :  x, 
AB  :  AL  : :  P+R  :  x'. 

In  like  manner  the  weight  T  of  the  whole  machine, 
supposed  to  be  united  at  the  centre  of  gravity  g,  may 
be  regarded  as  the  resultant  of  the  vertical  pressures 


102 


STATICS. 


•which  it  produces  upon  the  two  points  of  support :  and 
we  will  find  these  pressures  by  dividing  the  weight  T 
into  two  parts  reciprocally  proportional  to  the  distances 

A#>  #B- 

Then  let  Y  and  Y'  be  the  pressures  which  result  re 
spectively  at  the  points  A  and  B;  we  will  find  these 
pressures  by  the  two  proportions  which  follow : 

AB  :  B#  : :  T  :  Y, 
AB  :  Ag  : :  T  :  Y'. 

Lastly,  the  horizontal  force  s,  applied  to  the  point  c 
of  the  axis,  produces  horizontal  pressures  upon  the  two 
supports,  directed  perpendicularly  to  the  axis  AB,  and 
of  which  it  is  the  resultant :  we  will  find  likewise  these 
pressures  by  dividing  the  force  s  into  parts  reciprocally 
proportional  to  the  lines  AC,  CB.  Then  let  z,  z'  be  the 
horizontal  pressures,  produced  respectively  upon  the 


WHEEL   AND    AXLE.  163 

points  A,  B  :  we  will  find  these  pressures  by  the  pro 
portions  : 

AB  :  BC  : :  S  :  z, 
AB  :  AC  : :  S  :  z'. 

Thus  the  point  of  support  A  sustains  the  two  vertical 
pressures  x,  Y,  and  the  horizontal  pressure  which  acts 
in  the  direction  of  the  force  S.  In  like  manner  the  point 
B  sustains  the  vertical  pressures  x',  Y',  and  the  horizon 
tal  pressure  z'.  Hence,  by  compounding,  for  each  of 
these  points,  the  forces  which  act  upon  it,  we  will  find 
the  intensity  and  direction  of  their  resultant ;  and  we 
shall  have  the  intensity  of  the  resistance  of  which  it 
must  be  capable,  as  well  as  the  direction  in  which  it 
must  resist,  so  as  not  to  yield  to  the  united  efforts  of  the 
resistance  P,  the  power  Q,  and  the  weight  T  of  the  ma 
chine  ;  which  is  the  object  of  the  second  question. 


IV. 


166.  Hitherto  we  have  regarded  the  cords  as  infinitely 
fine  threads ;  but,  when  the  weight  P  is  suspended  by 
the  cord  KP,  the  line  of  direction  of  this  weight  does 
not  coincide  with  the  axis  of  the  cord ;  and  in  the  case 
where  the  cord  by  wrapping  around  the  cylinder  does 
not  change  in  figure,  its  axis  is  always  at  a  distance 
from  the  surface  of  the  cylinder  equal  to  the  semi-di 
ameter  of  the  cord.  Thus,  by  reason  of  its  thickness, 
the  cord  is  in  the  same  condition  as  though,  being  in 
finitely  fine  and  reduced  to  its  axis,  it  were  wrapped 


164 


STATICS. 


upon  a  cylinder  whose  radius  was  greater  than  the  radius 
of  the  cylinder  of  the  machine,  by  a  quantity  equal  to 
the  semi-diameter  of  the  cord.  It  is  the  same  with  the 
cord  of  the  wheel,  -which,  by  reason  of  its  thickness, 
may  be  regarded  as  a  mathematical  line  wrapped  upon 
a  wheel  whose  radius  is  greater  than  that  of  the  wheel 
of  the  machine,  by  a  quantity  equal  to  the  semi-diameter 
of  this  cord.  Hence,  in  all  the  relations  we  have  just 
found,  it  is  necessary  to  augment  the  radius  of  the 
cylinder  and  that  of  the  wheel  by  quantities  respectively 
equal  to  the  semi-diameters  of  the  cords  which  envelope 
them.  Thus,  for  example,  in  the  case  of  equilibrium  of 
the  wheel  and  axle,  the  power  Q  is  to  the  resistance  P  as 
the  radius  of  the  cylinder,  augmented  by  the  radius  of 
the  cord  KP,  is  to  the  radius  of  the  wheel,  augmented  by 
the  radius  of  the  cord  DQ. 


Fig.  70. 


V. 


167.  If  several  wheels  and  axles 
are  so  arranged  that  the  cord  BQ  of 
the  first  wheel,  being  drawn  by  a 
power  Q,  the  cord  CB  of  the  cylin 
der  of  this  wheel,  instead  of  being  at 
tached  immediately  to  the  resistance, 
is  wound  around  the  second  wheel ; 
the  cord  FH  of  the  cylinder  of  the 
second  is  likewise  wound  around  the 
wheel  of  the  third ;  and  lastly,  the  cord 
IP  of  the  cylinder  of  the  last  is  applied 
to  the  resistance  P  ;  the  power  and  the 


WHEEL   AND   AXLE.  165 

resistance  will  not  be  in  equilibrium,  unless  there  is  also 
equilibrium  between  the  two  forces  which  act  upon  each 
separate  wheel.  Thus,  by  naming  K  the  tension  of  the 
cord  CE,  and  L  that  of  the  cord  FH,  in  the  case  of 
general  equilibrium  we  have :  1st,  by  reason  of  the 
equilibrium  of  the  first  wheel  and  axle  (164), 

Q  :  K  : :  CA  :  AB  ; 
2d,  by  reason  of  the  equilibrium  of  the  second, 

K  :  L  : :  DF  :  DE  ; 
3d,  by  reason  of  the  equilibrium  of  the  third, 

L  :  P  : :  GI  :  GH  ; 

and  so  on,  whatever  may  be  the  number  of  machines. 
Hence,  by  multiplying  all  these  proportions  in  order,  we 
have 

Q  :  p  : :  CAXDFXGI  :  ABXDEXGH  ; 

that  is  to  say,  the  power  is  to  the  resistance  as  the  pro 
duct  of  the  radii  of  the  cylinders  is  to  the  product  of 
the  radii  of  the  wheels. 

For  example,  if  the  radius  of  the  wheel  of  each  is 
four  times  the  radius  of  its  cylinder,  in  the  case  of 
equilibrium  of  all  three  wheels  and  axles,  we  have 

Q  :  p  : :  Ixlxl  :  4x4x4,  or  : :  1  :  64. 


166 


STATICS. 


Hence  we  see,  that  by  multiplying  in  this  manner  the 
number  of  machines  we  can  put  moderate  powers  in  a 
state  of  equilibrium  with  very  great  resistances;  but 
this  arrangement  is  almost  never  employed,  because  it 
requires  too  great  a  length  of  cord  in  the  first  wheels, 
while  the  space  which  the  resistance  has  to  traverse  is 
inconsiderable. 


VI. 


168.  When  we  wish  to  profit  by 
the  advantages  of  this  arrange 
ment,  1st,  the  cords  are  suppressed 
which  transmit  the  motion  from  one 
wheel  and  axle  to  another ;  2d,  on 
the  circumference  of  each  wheel, 
teeth  are  placed  equally  distant 
from  each  other;  3d,  on  the  ar 
bor  of  each  of  the  toothed-wheels, 
another  wheel  similarly  toothed, 
of  a  smaller  diameter,  and  which  is 
called  the  pinion,  is  fixed  solidly;  4th,  lastly,  the 
whole  system  is  so  arranged  that  the  teeth  of  each 
pinion  interlock  with  the  teeth  of  the  following  wheel. 
By  this  means  one  wheel  cannot  turn  upon  its  axis,  with 
out  its  pinion  communicating  motion  to  the  wheel  with 
which  its  teeth  are  engaged,  and  causing  it  to  turn  upon 
its  axis;  and  the  number  of  revolutions  which  each 
pinion  can  cause  the  following  wheel  to  make,  is  un 
limited.  The  system  is  then  in  the  same  condition  as 
though  the  pinion,  considered  as  the  cylinder  of  a  wheel 


COG   WHEELS.  167 

and  axle,  and  the  wheel  with  which  it  interlocks,  were 
embraced  by  the  same  cord  as  in  the  preceding  case. 

Hence,  when  a  power  Q,  applied  to  the  circumference 
of  the  first  wheel,  is  in  equilibrium  with  a  resistance  P, 
applied  to  the  circumference  of  the  last  pinion,  the 
power  is  to  the  resistance  as  the  product  of  the  radii 
of  the  pinions  is  to  the  product  of  the  radii  of  the 
wheels. 


VII. 


169.  Toothed  wheels  are  employed  in  a  very  great 
number  of  machines,  principally  in  mills  and  in  the 
works  of  time  pieces.  Their  immediate  object  is  to 
communicate  to  a  cylinder  or  arbor  a  movement  of  ro 
tation  about  its  axis,  by  the  aid  of  the  rotary  motion  of 
another  arbor.  For  this  purpose  it  is  not  necessary  that 
the  axes  of  the  two  arbors  should  be  parallel,  as  we 
have  heretofore  supposed ;  it  is  sufficient  that  they  are 
in  the  same  plane. 

Fig.  72.  170.  When  the  axes  of  the  two  arbors 

are  at  right  angles,  the  teeth  ordinarily 
are  placed  perpendicularly  to  the  plane 
of  the  wheel,  as  may  be  seen  in  Fig.  72  ; 
then  they  can  interlock  with  those  of  the 
pinion,  or  with  the  staves  of  the  trundle 
AB,  which  produces  the  effect  of  a  pinion : 
by  this  interlocking,  the  teeth  of  the 
wheel  are  forced  to  slide  upon  the  staves 
in  the  direction  of  the  axis  of  the  trundle,  which  also 
causes  friction. 


168 


STATICS. 


171.  In  general,  whatever 
may  be  the  angle  BAG,  which  the 
axes  of  the  two  arbors  make, 
%  so  that  the  movement  of  rota 
tion  of  the  one  is  communicated 
to  the  other  by  means  of  two 
toothed  wheels  DEFG,  EHIF,  and 
the  teeth  do  not  slide  upon  each 
other  in  the  direction  of  the  axes,  it  is  necessary  that 
these  two  wheels,  called  leveled  wheels,  should  be  trunca 
ted  sections  of  two  cones  DAE,  EAH,  which  have  the 
same  summit  A,  and  whose  axes  coincide  with  those  of 
the  arbors ;  moreover,  the  teeth  of  the  two  wheels  should 
be  terminated  by  conic  surfaces  which  have  for  their 
summit  the  common  point  A. 


VIII. 

172.  The  jack-screw  is  also  a  machine  which  may  be 
referred  to  the  wheel  and  axle. 

The  simple  jack-screw  is  composed  of 
a  bar  of  iron  AB,  furnished  with  teeth 
upon  one  of  its  sides,  and  moveable  in 
the  direction  of  its  length  within  a  box 
DE.  The  teeth  of  the  bar  interlock  with 
those  of  a  pinion  c,  which  is  turned 
upon  an  axis  by  means  of  a  crank  F. 
The  teeth  of  the  pinion  carry  along  those 
of  the  bar,  and  cause  the  weight  to  rise,  which  rests 
upon  the  head  A  of  the  bar,  or  which  is  raised  by  the 
hook  B.  This  machine,  evidently,  is  nothing  else  than 
a  wheel  and  axle,  and,  in  the  case  of  equilibrium,  by 


THE   INCLINED    PLANE.  169 

supposing  the  direction  of  the  power  to  be  perpendicu 
lar  to  the  arm  of  the  crank,  the  power  is  to  the  resist 
ance  as  the  radius  of  the  pinion  is  to  the  arm  of  the 
crank. 

In  the  compound  jack-screw,  the  teeth  of  the  first 
pinion  interlock  with  those  of  a  second  toothed  wheel, 
and  the  teeth  of  the  pinion  of  this  wheel  interlock  with 
those  of  the  bar.  By  this  means  the  power  is  placed  in 
a  state  of  equilibrium  with  a  greater  resistance.  This 
case  is  related  to  that  of  toothed  wheels,  and  we  will 
not  dilate  any  more  upon  that  subject. 

173.  When  two  powers  are  in  equilibrium  by  means 
of  a  pulley  or  wheel  and  axle,  it  is  very  evident  that 
we  may  consider  them  as  though  they  were  in  equili 
brium  at  the  extremities  of  a  lever  whose  fulcrum  is  in 
the  axis  of  the  cylinder  or  the  pulley ;  hence  (151), 
these  powers  are  to  each  other  reciprocally,  as  the 
spaces  which  they  would  traverse  along  their  directions, 
if  the  equilibrium  were  disturbed  infinitesimally. 


ARTICLE  III. 
On  the  Equilibrium  of  the  Inclined  Plane. 


174.  A  plane  is  said  to  be  inclined, 
when  it  makes  an  angle  with  a  horizon 
tal  plane  ABEF,  and  this  angle  is  not  a 
right  angle. 

175.  If  through  any  point  H,  taken 
upon  the   line    of    intersection   AB    of 
the  inclined  plane  and  the  horizontal 


170  STATICS. 

plane,  two  perpendiculars  be  drawn  to  this  line,  the  one 
HK  in  the  horizontal  plane,  and  the  other  HI  in  the  in 
clined  plane,  the  angle  IHK,  formed  by  these  two  per 
pendiculars,  is  the  measure  of  the  inclination  of  the 
plane.  The  plane  of  the  angle  IHK,  which  passes 
through  two  lines  perpendicular  to  AB,  is  perpendicular 
to  the  line  AB  ;  hence  it  is  perpendicular  to  the  two 
planes  ABCD  and  ABEF,  which  intersect  in  this  line; 
hence  this  plane  is  at  the  same  time  vertical  and  per 
pendicular  to  the  inclined  plane. 

176.  Reciprocally,  every  plane  which  is  at  the  same 
time  vertical  and  perpendicular  to  the  inclined  plane,  is 
perpendicular  to  the  intersection  AB  of  the  inclined 
plane  with  the  horizontal  plane  ;  hence,  the  lines  HI  and 
HK  along  which  it  intersects  these  two  last  planes,  are 
also  perpendicular  to  AB,  and  form  between  them  an 
angle  IHK,  which  is  the  measure  of  the  inclination  of  the 
plane  ABCD  with  reference  to  the  horizontal  plane. 

177.  If  through  the  point  I  there  be  drawn  a  vertical 
line  IL,  this  line  will  not  leave  the  plane  of  the  angle 
IHK  ;  it  will  meet  the  horizontal  line  HK  to  which  it  will 
be  perpendicular,  and  it  will  form  a  rectangular  triangle 
IHL.     The    hypotenuse   of  this  triangle  is  named  the 
length  of  the  inclined  plane ;  the  side  IL  is  its  height, 
and  the  other  side  HL  is  its  base. 


178.  When  a  body,  which,  in  a  single  point  Q,  touches 
an  immoveable  plane  ABCD,  is  pushed  by  a  single  force 
p,  whose  direction  PQ,  1st,  passes  through  the  point  of 


THE   INCLINED   PLANE. 


171 


Fig.  76.  contact  Q,  2d,  is  perpendicular 

to  the  plane,  this  body  remains 
at  rest. 

Thus,  we  can  conceive  the 
force  P  to  be  applied  to  the  point 
Q  of  its  direction.  Now  this  di 
rection  being  perpendicular  to 
the  plane,  and  consequently  to 
all  the  lines  QR,  QS,  QT,  which  we  can  draw  in  the  plane 
through  the  point  Q,  it  is  similarly  disposed  with 
reference  to  all  these  lines  ;  there  is  then  no  reason  why 
the  point  Q  should  move  along  one  of  the  lines  rather 
than  along  any  other;  hence  this  point,  and  conse 
quently  the  whole  body,  will  remain  at  rest. 

Fig.  77.  179.  Both  the  conditions  just  men 

tioned  are  necessary  to  the  rest  of  the 
body :  for,  1st,  if  the  direction  n 
of  the  force  does  not  pass  through 
the  point  of  contact,  nothing  prevents 
the  point  a  of  the  body  which  is  upon 
this  direction  from  approaching  the 
plane,  and  the  body  would  be  put  in  motion.  2d.  If 
the  direction  PQ  of  the  force  passes  through  the  point 
Fig.  78.  Of  contact,  and  is  not  perpendicular 

to  the  plane,  by  conceiving  this  force 
to  be  still  applied  to  the  point  Q,  let 
its  direction  be  prolonged  to  QR,  and 
through  the  point  Q  draw  the  line  QS 
perpendicular  to  the  plane  ;  through 
the  two  lines  QR,  QS  draw  a  plane, 
which  will  intersect  the  first  plane 
ABCD  somewhere  in  a  line  Hi,  which  will  pass  through 


p 

a 


172  STATICS. 

the  point  Q.  This  being  done,  if  we  represent  the  force 
p  by  the  part  QR  of  its  direction,  and  complete  the  pa 
rallelogram  QTRS,  instead  of  the  force  P  we  may  take 
the  two  forces  represented  by  QS  and  QT.  Now  the 
force  QS,  which  passes  through  the  point  of  contact,  and 
whose  direction  is  perpendicular  to  the  plane  ABCD,  will 
be  destroyed  by  the  resistance  of  this  plane  (178) ;  but 
nothing  will  oppose  the  action  of  the  force  QT,  whose 
direction  is  parallel  to  the  plane ;  the  point  Q  will  then 
move  in  the  direction  QII,  and  the  body  will  not  be  at 
rest. 

Fig.  78.  180.  Hence  it  follows,  that,  when  a 

body,  affected  by  the  simple  action 
x"'\  of  its  gravity,  remains  in  equilibrium 

upon  a  plane  ABCD,  which  it  touches 
in  a  single  point  Q,  1st,  the  centre 
of  gravity  P  of  this  body,  and  the 
point  °f  contact  Q,  are  in  the  same 
A~~  vertical  line;  2d,  the  plane  ABCD  is 

horizontal;  for,  since  the  weight  of  the  body  can  be 
regarded  as  a  single  force  applied  to  its  centre  of  gravity, 
the  body  cannot  be  at  rest,  unless  the  direction  of  this 
force  passes  through  the  point  of  contact  and  is  perpen 
dicular  to  the  plane  upon  which  the  body  is  sustained. 
Fig.  79.  181.  It  follows  again,  that 

when  a  body,  affected  by  the 
single  action  of  gravitation,  is 
sustained  upon  an  inclined  plane 
ABCD  by  a  single  point  Q,  and 
this  point  is  found  in  the  verti 
cal  drawn  through  the  centre  of  gravity,  this  body  must 
tend  to  slip  upon  the  plane  ;  and  the  direction  QH  along 


THE   INCLINED    PLANE. 


173 


Fig.  80. 


which  the  point  Q  tends  to  move,  is  the  intersection  of 
the  plane  ABCD  with  the  plane  IHK,  which  is  at  the 
same  time  vertical  and  perpendicular  to  the  inclined 
plane. 

182.  What  has  just 
been  said  of  a  body 
pushed  by  a  single  force 
against  a  plane,  must  ap 
ply  to  a  body  pushed 
against  a  curved  surface 
AQB,  which  it  touches  only 
in  the  single  point  Q :  that 
is  to  say,  this  body  cannot 
be  at  rest,  unless  the  direction  of  the  force  which  pushes 
it  passes  through  the  point  Q,  and  is  perpendicular  to 
the  curved  surface  at  this  point ;  for  this  body  may  be 
considered  as  sustained  upon  the  plane  DE  tangent  to 
the  curved  surface  at  the  point  Q. 

183.  We  see,  then,  that  when  a  lever  may  slide  upon 
its  fulcrum,  it  is  not  sufficient  for  its  remaining  at  rest, 
that  the  resultant  of  the  two  powers  which  are  applied 
to  this  lever  are  directed  towards  the  fulcrum ;  it  is  be 
sides  necessary  that  the  direction  of  this  resultant 
should  be  perpendicular  to  the  surface  of  the  lever  at 
the  point  where  it  touches  the  fulcrum. 


II. 

184.  When  a  body,  pushed  by  a  single  force  P 
against  an  immoveable  plane  ABCD,  is  sustained  upon 
this  plane  by  a  definite  base  VXYZ,  if  the  direction  PQ 

15* 


174  STATICS. 

of  the  force  meets  the  base  somewhere  in  a  point  Q,  and 
if  it  is  at  the  same  time  perpendicular  to  the  plane,  the 
body  is  at  rest ;  for  we  have  seen  (178)  that  if  the  base 
Fi9-  81.  were  reduced  to  the  single 

point  Q,  rest  would  take  place  ; 
it  is  evident  that  the  other 
points  of  support,  which  the 
base  presents,  cannot  disturb 
it. 

We  will  demonstrate,  as  in 
No.  179,  that  both  these  con 
ditions  are  necessary  to  the 
body's  remaining  at  rest :  the  effect  of  the  first  is  to 
prevent  the  body  from  turning  upon  one  of  the  sides  of 
its  base ;  the  effect  of  the  second  is  to  prevent  it  from 
slipping  upon  the  plane  ABCD. 

185.  If  the  body,  instead  of  being  supported  upon 
the  plane  by  a  continuous  base,  simply  touches  it  by 
several  points  separated  from  each  other,  we  may  regard 
these  points  as  the  summits  of  the  angles  of  a  polygonal 
base ;  and  the  body  is  at  rest  when  the  direction  of  the 
force,  which  pushes  it  against  the  plane,  is  perpendicu 
lar  to  this  plane,  and  at  the  same  time  passes  through 
the  interior  of  the  polygon. 

Thus,  since  the  weight  of  a  body  may  be  regarded  as 
a  vertical  force  applied  to  its  centre  of  gravity  P,  we 
see,  1st,  that  a  body,  which  rests  with  its  base  upon  a 
horizontal  plane  ABCD,  cannot  be  stable,  unless  the  ver 
tical  PQ,  drawn  through  the  centre  of  gravity,  meets 
some  point  Q  of  the  base ;  2d,  if  the  body  rests  upon 
the  plane  by  a  certain  number  of  points  of  support, 


c 


-   U|  ' 


THE    INCLINED    PLANE.  175 


u,  x,  Y,  .  .  .  it  cannot  be  stable, 
unless  the  vertical  PQ,  drawn 
through  its  centre  of  gravity 
F,  passes  through  a  point 
Q  taken  in  the  interior  of  the 
polygon  UXY,  which  may  be 
formed  by  joining  the  exterior 
points  of  support  by  the  lines 
ux,  XY,  YU. 


III. 

186.  Hitherto  we  have  supposed  that  the  body,  sup 
ported  upon  a  plane,  was  pushed  by  a  single  force  ;  but  it 
is  evident  that,  if  the  body  is  pushed  by  several  forces 
at  the  same  time,  it  cannot  be  at  rest,  unless  the  result 
ant  of  all  these  forces  satisfies  the  preceding  conditions  ; 
that  is  to  say,  unless  the  direction  of  this  resultant  is 
perpendicular  to  the  plane,  and  passes  through  the  base 
of  the  body.  There  is  then  in  this  case  a  third  con 
dition  necessary  to  the  rest  of  the  body,  and  this  condi 
tion  is,  that  all  the  forces  which  act  upon  it  must  have 
one  resultant. 

The  whole  theory  of  the  equilibrium  of  a  body,  pushed 
by  as  many  forces  as  we  please  to  assume,  and  supported 
upon  a  single  resisting  plane,  consists  in  the  search  for 
the  directions  and  intensities  which  the  forces  must  have, 
so  that  the  three  conditions  just  mentioned  may  be  ful 
filled.  We  will  content  ourselves  with  developing  it  for 
a  few  simple  cases,  and  principally  for  that  in  which  the 
body  is  pushed  by  two  forces. 


176 


STATICS. 


IV. 

Fi9-  81.  187.  Let  LUZ  be  a  body  sup 

ported  by  a  base  UXYZ  upon  a 
resisting  plane  ABCD,  and  so 
licited  at  the  same  time  by  two 
forces  R,  s.  According  to  the 
preceding  remarks,  in  order 
that  the  body  may  be  at  rest, 
it  is  necessary,  1st,  that  the 
two  forces  R,  s  should  have 
one  resultant :  now,  two  forces  cannot  have  one  result 
ant,  unless  their  directions  are  in  the  same  plane,  (10) ; 
hence,  1st,  the  directions  of  the  two  forces  R,  S  must  be 
included  in  the  same  plane,  and  intersect  in  a  certain 
point  N. 

2d.  It  is  necessary  that  the  direction  PNQ  of  the  re 
sultant  of  the  two  forces  R,  s,  should  be  perpendicular 
to  the  plane  ABCD  :  now  the  resultant  of  the  two  forces 
is  always  contained  in  the  plane  drawn  through  their 
directions ;  hence,  2d,  the  plane  in  which  the  two  forces 
R,  s  are  directed,  should  be  perpendicular  to  the  plane 
ABCD. 

3d.  It  is  necessary  that  the  direction  of  the  resultant 
should  pass  through  a  point  Q  of  the  base. 

188.  From  this  it  follows,  that  if  one  of  the  two 
forces,  for  example  the  force  R,  is  the  weight  of  the 
body  which  may  be  considered  as  applied  to  the  centre 
of  gravity  M,  and  whose  direction  MR  is  vertical,  the 
body  cannot  be  at  rest  upon  an  inclined  plane  ABCD, 
unless  the  direction  NS  of  the  other  force  is  contained 


THE   INCLINED   PLANE. 


177 


in  a  vertical  plane,  drawn  through  the  centre  of  gravity 
M,  perpendicularly  to  the  inclined  plane,  and,  moreover, 
that  the  direction  PQ  of  the  resultant  of  the  two  forces 
is  perpendicular  to  the  inclined  plane,  and  passes  through 
a  point  Q  of  the  base  of  the  body. 

Now  all  these  conditions,  relative  to  the  directions  of 
the  forces,  being  supposed  to  be  filled,  we  will  seek  the 
relations  which  the  two  forces  E,  s,  and  the  weight  P 
of  the  plane,  have  for  each  other  in  the  case  of  equi 
librium. 


Fig.  83. 


V. 

189.  Let  LXU  be  a  body  sup 
ported  by  its  base  ux  upon  a  re 
sisting  plane  HI,  and  kept  at  rest 
upon  this  plane  by  the  two  forces 
E,  S.  Having  prolonged  the  di 
rections  of  the  two  forces  until 
they  meet  in  a  point  N,  draw 
B.  through  this  point  the  line  NP  per 

pendicular  to  the  plane  Hi.  We  have  seen  that  this 
line  will  be  the  direction  of  the  resultant  of  the  two 
forces  E,  s.  Hence,  if  we  represent  this  resultant  by 
the  part  NE  of  its  direction,  and  if,  by  drawing  through 
the  point  E  the  lines  EG,  EF,  parallel  to  the  directions 
of  the  forces  E,  s,  we  complete  the  parallelogram  NFEG, 
the  sides  NF,  NG  will  represent  the  intensities  of  the 
forces  E,  s.  Hence,  by  naming  P  the  weight  on  the 
plane  which  is  equal  to  the  resultant,  we  shall  have 


E  :  s  :  P  : :  NF  :  NG  or  FE  :  NE. 


178  STATICS. 

In  order  to  have  the  ratios  of  these  three  forces  ex 
pressed  in  quantities  independent  of  the  construction 
of  the  parallelogram  NFEG-,  we  will  remark,  that  in  the 
triangle  NEF  the  sides  are  to  each  other  in  the  ratio  of 
the  sines  of  the  opposite  angles,  or  we  shall  have 

NF  :  FE  :  EN  : :  sin  NEF  :  sin  FNE  :  sin  NFE. 
Hence  we  shall  have 

E  :  s  :  P  : :  sin  NEF  :  sin  FNE  :  sin  NFE. 

Now  these  three  angles  are  those  which  form  with  each 
other  the  directions  of  the  three  forces  E,  •  s,  P  ;  hence 
these  forces  are  to  each  other  as  the  sine  of  the  angle 
which  the  directions  of  the  other  two  forms. 

We  see  then  that,  of  the  six  things 
which   may   be   considered   in    this 
equilibrium,  and  which  are,  the  di 
rections  of  the  three  forces  and  their 
intensities,  any  three  being  given,  we 
can  determine  the  other  three,  in  all 
cases  where,  of  the  six  things  which 
we  consider  in  the  triangle  NEF,  namely,  the  angles  and 
the  sides,  the  three  analogous  ones  being  given,  we  can 
determine  the  three  others. 

190.  If  one  of  the  forces,  for  example  the  force  E, 
is  the  weight  of  the  body  whose  direction  is  vertical, 
and  passes  through  the  centre  of  gravity  M,  and  the  di 
rection  of  the  force  S,  which  retains  the  body  in  equi 
librium  upon  the  inclined  plane,  is  parallel  to  this  plane, 
draw  the  base  KH  and  the  height  KI  of  the  inclined 


THE   INCLINED    PLANE. 


179 


plane ;  the  right-angled  triangles  NFE,  IHK  will  be  simi 
lar,  because  the  angles  NFE,  HIK,  whose  sides  are  paral 
lel  each  to  each,  will  be  equal,  and  we  shall  have 

NF  :  FE  :  EN  : :  HI  :  IK  :  KH  ; 
hence  we  shall  also  have 

R  :  s  :  P  : :  HI  :  IK  :  KH. 

Thus,  in  this  case,  the  weight  of  the  body  is  to  the 
force  which  holds  it  in  equilibrium,  as  the  length  of  the 
inclined  plane  is  to  its  height. 

Fi9-  85-  191.  By  supposing  still  that 

s  the  force  R  is  the  weight  of  the 
body,  if  the  direction  of  the 
force  S  is  horizontal,  and  con 
sequently  parallel  to  the  base 
HK  of  the  inclined  plane,  the 
right-angled  triangles  NFE,  HKI  are  again  similar ;  be 
cause  the  sides  of  the  one  will  be  perpendicular  to  the 
sides  of  the  other,  and  we  shall  have 


NF  :  FE  :  EN  : :  HK  :  KI  :  IH, 


we  shall  also  have 


R  :  s  :  P 


HK  :  KI  :  m. 


Hence,  the  weight  of  the  body  is  to  the  force  which 
holds  it  in  equilibrium,  as  the  base  of  the  inclined  plane 
is  to  its  height. 


180 


STATICS. 


VI. 


192.  Let  us  now  consider 
the  equilibrium  of  a  body  sup 
ported  by  two  inclined  planes. 
Let  M  be  a  body  subjected  to  the 
single  action  of  gravity,  and  re 
tained  by  the  two  inclined  planes 
ABCD,  ABEF,  which  intersect  some 
where  in  the  line  AB.  Let  H,  I  be  the  points  at  which 
the  body  touches  the  two  planes,  and  GR  the  vertical 
line  drawn  through  its  centre  of  gravity,  and  which  will 
consequently  be  the  direction  of  its  weight.  It  is  evi 
dent  that  this  body  cannot  be  at  rest,  unless  its  weight 
R  can  be  decomposed  into  two  other  forces  P,  Q,  applied 
to  the  same  body,  and  which  are  destroyed  by  the  re 
sistance  of  the  two  planes :  or,  what  is  the  same  thing, 
unless  the  directions  of  the  two  forces  P,  Q  pass  through 
the  points  of  support  I,  H,  and  are  perpendicular  each 
to  the  corresponding  inclined  plane.  Now  the  direction 
of  a  force  and  those  of  its  two  components  are  contained 
in  the  same  plane,  and  necessarily  meet  in  the  same 
point ;  hence,  in  order  that  the  body  M  may  be  at  rest 
between  the  two  inclined  planes,  it  is  necessary  that  the 
perpendiculars  IG,  HG,  drawn  through  the  points  of  sup 
port  I,  H  to  the  two  inclined  planes,  should  be  in  the 
same  plane  with  the  vertical  drawn  through  the  centre 
of  gravity  of  the  body,  and  intersect  this  vertical  in  the 
same  point  G. 


THE    INCLINED    PLANE.  181 

193.  Hence  it  follows,  that,  in  order  that  a  body  M 
may  be  at  rest  between  two  inclined  planes,  inde 
pendently  of  the  position  of  the  body,  the  planes  should 
satisfy  the  condition  that  the  line  AB  of  their  intersec 
tion  should  be  horizontal.  Thus,  the  plane  IGH,  which 
must  contain  the  vertical  GR,  and  the  perpendiculars 
IG,  HG  to  the  two  inclined  planes,  is  at  the  same  time 
vertical  and  perpendicular  to  these  two  planes ;  hence, 
reciprocally,  the  two  inclined  planes  must  be  perpen 
dicular  to  the  vertical  plane  IGH ;  hence  the  line  AB 
of  their  intersection  should  be  perpendicular  to  this 
same  plane,  and  consequently  horizontal. 

87.  194.    These  conditions, 

which  have  for  object  the 
respective  positions  of  the 
body  and  of  the  two  in 
clined  planes,  being  sup 
posed  to  be  fulfilled,  in 
order  to  find  the  ratio  of 
the  weight  R  of  the  body 
to  the  weights  P,  Q,  which  the  two  planes  support,  we 
will  remark  that  the  plane  IGH,  vertical  and  perpendicu 
lar  to  the  two  inclined  planes  containing  the  angles 
which  these  two  planes  form  with  the  horizon,  includes 
all  that  relates  to  the  question,  and  we  may  be  content 
to  consider  it  alone,  as  in  Fig.  87.  Hence,  let  AD,  AF 
be  the  intersections  of  the  vertical  plane  IGH  with  the 
two  inclined  planes  ;  these  lines  form  with  the  horizon 
tal  line  vz,  or  with  any  other  horizontal  line  XY,  angles 
which  will  measure  the  inclinations  of  the  two  planes. 
This  being  granted,  if  we  represent  the  weight  of  the 


16 


182  STATICS. 

body  by  the  part  GR  of  its  direction,  and  complete  the 
parallelogram  GPQR,  we  shall  have 

R  :  p  :  Q  : :  GR  :  RQ  :  QG. 

Now  the  triangles  GQR,  XAY,  whose  sides  are  perpendicu 
lar  each  to  each,  give 

GR  :  RQ  :  QG  : :  XY  :  YA  :  AX  ; 
hence  we  shall  also  have 

R  :  p  :  Q  : :  XY  :  YA  :  AX  ; 

or,  lastly,  because  the  sides  of  the  triangle  XAY  are  pro 
portional  to  the  sines  of  the  opposite  angles,  we  shall 
have 

R  :  p  :  Q  : :  sin  YAX  :  sin  AXY  :  sin  XYA, 

that  is  to  say,  by  representing  the  weight  of  the  body 
by  the  sine  of  the  angle  which  the  two  inclined  planes 
form  with  each  other,  the  weights  which  these  two 
planes  support  are  to  each  other  reciprocally  as  the 
sines  of  the  angles  which  these  planes  form  with  the 
horizon. 


THE   INCLINED   PLANE.  183 


VII. 

195.  Lastly,  if  a  body  is 
sustained  by  three  points 
A,  B,  c,  upon  three  inclined 
planes,  it  is  evident  that 
this  body  cannot  be  at  rest 
unless  its  weight  p,  whose 
direction  DP  is  vertical,  and 
passes  through  the  centre  of  gravity  of  the  body,  can 
be  decomposed  into  three  other  forces  Q,  E,  s,  which  are 
destroyed  by  the  resistances  of  the  inclined  planesj 
that  is  to  say,  unless  the  directions  of  the  three  forces 
Q,  n,  s  pass  through  the  three  points  of  support,  and 
are  each  perpendicular  to  the  corresponding  inclined 
plane. 

196.  In  order  that  the  body  may  be  at  rest,  it  is 
necessary  that  the  weight  P  should  be  susceptible  of 
being  decomposed  into  two  forces  Q,  x;  the  first  of 
which,  Q,  being  directed  towards  one  of  the  points  of 
support  A,  perpendicularly  to  the  inclined  plane  which 
passes  through  this  point,  the  other  force  x  may  itself 
be  decomposed  into  two  others  B,  s,  in  the  directions 
of  the  two  other  points  of  support  B,  c,  perpendicular 
to  the  other  inclined  planes. 

Hence  we  see,  in  this  case,  it  is  not  necessary  that 
the  perpendiculars  to  the  inclined  planes,  drawn  through 
the  points  of  contact  A,  B,  c,  should  all  three  meet  in 
the  same  point,  nor  even  that  they  should  all  three  meet 
the  vertical  drawn  through  the  centre  of  gravity  of  the 
body. 


184  STATICS. 


THEOREM. 

197.  When  a  body  without  gravity,  supported  upon 
an  inclined  plane  EC  by  a  single  point  C,  is  in  equili 
brium  between  two  poivers  P,  Q,  these  powers  are  to  each 
other  reciprocally  as  the  spaces  which  they  ivould  tra 
verse  in  the  line  of  their  directions,  if  the  equilibrium 
were  disturbed  infinitesimally. 

Fig-  89.  DEMONSTRATION.     The  two 

-g  powers  P,  Q  being  in  equili 
brium,  their  resultant  should 
B  be  perpendicular  to  the  inclined 
plane,  and  pass  through  the 
point  of  support  c  (186) ;  it 
follows  from  this  that  the  di- 
rections  of  these  powers  should 
coincide  in  a  certain  point  G,  and  that  the  line  GO 
should  be  perpendicular  to  the  inclined  plane.  This 
being  established,  from  the  point  c  draw  upon  the  di 
rections  of  the  powers  P,  Q,  the  perpendiculars  CE,  OF  : 
it  is  evident  that  the  angle  EOF,  supposed  to  be  invaria 
ble,  may  be  considered  as  a  bent  lever,  at  the  extremi 
ties  of  which  are  applied  the  two  powers  in  equilibrium 
about  the  point  of  support  c :  hence  we  can  demonstrate, 
as  in  No.  151,  that  the  powers  are  to  each  other  recipro 
cally  as  the  spaces  which  they  would  traverse  in  the  line 
of  their  directions,  if  the  equilibrium  were  disturbed 
infinitesimally. 


THE   SCREW. 


185 


On  the  Screw. 

198.  If  we  conceive  a  cylin 
der  ABCD  to  be  enveloped  by 
a  thread  AGHIK  . .  . .,  and  so 
disposed  that  the  angles  FLO, 
FMP,  FNQ  .  .  .  .,  formed   by 
the  direction  of  the  thread 
with  the  lines  drawn  upon  the 
surface  of  the  cylinder  paral 
lel  to  the  axis,  are  equal  to  each  other,  the  curve  which 
the  thread  traces  upon  the  surface  of  the  cylinder  is 
named  a  helix. 

199.  Hence  it  follows,  that  if  we  develope  the  surface 
of  the  cylinder,  and  extend  it  upon  a  plane,  as  we  see 
in  abed,  1st,  the  developement  ah'  or  lik'  of  one  revolu 
tion  of  the  helix  will  be  a  straight  line ;  because  the 
angles  which  this  line  will  form  with  all  such  lines  as  ef, 
drawn  parallel  to  the  side  ad,  will  be  equal  to  eact 
other.  2d.  This  developement  ah'  of  a  revolution  of 
the  helix  will  be  the  hypotenuse  of  a  right-angled  tri 
angle  abh',  whose  base  ab  will  be  equal  to  the  circum 
ference  of  the  base  of  the  cylinder,  and  whose  height 
bh'  will  be  equal  to  the  distance  of  the  revolution  which 
we  consider,  from  the  revolution  which  'follows  it.  3d. 
All  the  hypotenuses  ah',  hJc'  being  parallel  to  each 
other,  the  right-angled  triangles  abh',  hh'k'  .  .  .,  will  be 
equal  and  similar,  and  will  have  equal  heights.  Hence 
the  intervals  LM,  MN,  .  . .  between  two  consecutive  revo 
lutions  of  the  helix,  considered  upon  the  surface  of  the 
cylinder,  are  everywhere  equal  to  each  other. 

16* 


186 


STATICS. 


Fig.  92. 


200.  This  being  established,  the  screw  may  be  con 
sidered  as  a  right  cylinder,  enveloped  by  a  projecting 
fillet,  adhering  and  wound  as  a  helix  upon  the  surface 
of  the  cylinder.     In  the  wooden  screw,  the  form  of  the 
fillet  is  such,  that  if  it  be  cut  by  a  plane  drawn  through 
the  axis  of  the  cylinder,  its  section  is  most  frequently 
an  isosceles  triangle,  as  may  be  seen  in  Fig.  92 ;  but 
in  large  iron  screws  which  are  made  with  care,  the  sec 
tion  of   the  fillet  is  rectangular,  as  in  Fig.  91.     The 
constant  interval  AB,  which  is  found  between  two  con 
secutive  revolutions  of  the  fillet,  is  named  the  pitch  of 
the  screw,  or  the  helical  interval. 

201.  The  piece  MN,  into  which  the  screw  enters,  is 
named  the  nut.    Its  cavity  is  invested  with  another  pro 
jecting  fillet,  adhering,  and  wound  likewise  as  a  helix ; 
and  whose  figure  is  such  that  it  exactly  fills  the  inter 
vals  left  by  the  fillets  of  the  screw.     Thus,  the  screw 
can  turn  in  its  nut,  but  it  cannot  do  so  without  moving 
in  the  direction  of  its  axis ;  and  for  one  entire  revolu 
tion,  it  moves  in  the  direction  of  the  axis  by  a  quantity 
equal  to  the  pitch  of  the  screw. 


THE    SCREW.  187 

202.  Sometimes   the  screw  is  fixed,  and  the  nut 
moves   around  it ;    then,  for  each  revolution,  the  nut 
is  carried  upon  the  screw  by  a  quantity  equal  to  the 
interval. 

203.  The  screw  may  serve  to  elevate  weights  or  over 
come  resistances ;  but   it  is  employed  most  generally 
when  great  pressures  are  proposed  to  be  exerted.     For 
this  purpose  we  apply  a  power  Q  to  the  extremity  of  a 
bar  which  traverses  the  head  of  the  screw,  Fig.  92,  or  the 
nut,  Fig.  91,  according  as  it  is  the  one  or  the  other  of 
these  two  pieces  which  is  moveable ;  and  this  power,  by 
causing  the  piece  to  turn  to  which  it  is  applied,  makes 
the  head  of  the  screw  advance  towards  the  nut,  or  re 
ciprocally.     The  objects  to  be  compressed  are  ranged 
between  two  plates ;  one  of  these  plates  is  fixed,  the 
other  is  pressed  by  the  movable  piece,  which  can  ad 
vance  only  by  reducing  the  volume  of  the  objects. 

We  now  propose,  disregarding  the  friction,  to  find  the 
ratio  of  the  power  Q  to  the  resistance  P,  which  is  in 
equilibrium  with  it  by  being  opposed  to  the  motion  of 
the  moveable  piece,  along  a  direction  parallel  to  the 
axis  of  the  screw ;  and,  because  the  effect  is  absolutely 
the  same,  whether  the  screw  turns  in  its  nut,  or  the  nut 
turns  upon  the  screw,  it  will  be  sufficient  to  examine  the 
latter  case. 

204.  The  screw  being  fixed  and  in  a  vertical  position, 
we  will  conceive  the  nut  to  be  left  to  the  action  of  gravi 
ty,  and  even,  if  we  please,  that  it  is  charged  with  an 

,-  additional  weight ;  it  is  evident  that  it  will  descend  by 
turning,  and  that  it  will  traverse  all  the  interior  fillets 
of  the  screw,  by  sliding  over  them  as  over  inclined 
surfaces.  It  is  also  evident  that  we  oppose  this  effect 


188 


STATICS. 


by  preventing  the  nut  from 
turning  around  the  screw,  and 
consequently  by  applying  to 
the  extremity  of  the  bar  FV  a 
power  Q  which  is  directed  per 
pendicularly  to  this  bar,  and 
in  a  plane  perpendicular  to 
the  axis  of  the  screw. 

205.  Suppose,  for  an  in 
stant,  that  the  nut  rests  upon 
the  surface  of  the  fillet  of  the  screw  only  in  a  single 
point ;  this  point,  during  the  motion  of  the  nut,  will  de 
scribe  a  helix,  whose  pitch  will  be  the  same  as  that  of  the 
screw,  and  which  we  may  conceive  to  be  traced  upon 
the  surface  of  a  cylinder,  whose  radius  would  be  equal 
to  the  distance  of  the  describing  point  from  the  axis  of 
the  screw.  Let  ABCD  be  the  cylinder,  EFGHI  the  helix 

in  question,  and  M  the 
point  of  the  nut  which 
describes  it.  Let  XY  be 
the  tangent  of  the  helix 
at  the  point  M ;  through 
a  point  Y  of  this  tangent 
draw  the  vertical  YZ,  equal 
to  the  pitch  of  the  helix,  and  in  the  vertical  plane  XYZ  of 
the  horizontal  line  xz  :  it  is  clear,  that  this  last  line  will 
be  equal  to  the  circumference  of  the  base  of  the  cylin 
der  ABCD  (199),  or  to  the  circumference  of  the  circle 
whose  radius  is  CK  ;  and  we  will  represent  it  by  circ.  CK. 
This  being  established,  the  point  M,  which  we  may 
.regard  as  charged  with  the  whole  weight  P  of  the  screw, 
will  be  supported  upon  the  helix  as  though  it  were  upon 


TIIE   SCREW.  189 

the  inclined  plane  XY  ;  thus,  in  order  to  hold  it  in  equi 
librium  by  means  of  a  force  R,  which  should  be  imme 
diately  applied  to  it,  and  which  should  be  directed  pa 
rallel  to  xz,  it  is  necessary  (191)  that  this  force  R  should 
be  to  the  weight  P,  as  the  height  of  the  inclined  plane 
is  to  its  base,  or  that  we  should  have, 

R  :  p  : :  YZ  :  circ.  CK. 

But  if,  instead  of  a  force  R  immediately  applied  to  the 
point  M,  we  employ  a  force  Q,  whose  direction  is  parallel 
to  that  of  the  first,  and  which  acts  at  the  extremity  of 
a  bar  M'/,  it  is  necessary  that  this  force  should  exert 
upon  the  point  M  the  same  effect  as  the  force  R,  and  for 
that  purpose,  these  forces  should  be  to  each  other  re 
ciprocally  as  their  distances  from  the  axis  of  the  cylin 
der  ;  that  is  to  say,  we  should  have, 

Q  :  R  : :  M/  :  M'/; 

or,  since  the  circumferences  of  circles  are  to  each  other 
as  their  radii,  we  should  have, 

Q  :  R  :  :  circ.  M/ :  circ.  m'f. 

Then,  by  multiplying  together  in  order  this  proportion 
and  the  first,  we  shall  have,  since  circ.  CK=circ.  M/, 

Q  :  p  : :  YZ  :  circ.  M'/; 

that  is  to  say,  the  power  which  retains  the  nut  in  equi 
librium,  will  be  to  the  weight  of  the  nut,  as  the  pitch  of 


190  STATICS. 

the  screw  is  to  the  circumference  of  the  circle  which  the 
power  tends  to  describe.* 

206.  Since  the  distance  of  the  point  M  from  the  axis 
of  the  screw  does  not  enter  into  this  proportion,  it  fol 
lows  that,  whatever  may  be  this  distance,  the  ratio  of 
the  weight  p  of  the  nut  to  the  power  Q,  which  is  in 
equilibrium  with  it,  is  always  the  same,  provided  this 
power  is  always  applied  to  the  same  point. 

207.  If  the  fillet  of  the  nut  is  supported  upon  that 
of  the  screw  by  several  points  unequally  distant  from 
the  axis  of  the  screw,  as  it  generally  happens,  the  total 
weight  of  the  nut  can  be  regarded  as  divided  into  par- 

tial  weights,  each  applied  to 
one  of  the  points  of  support. 
Now  the  partial  power  ap 
plied  to  the  point  v,  and  which 
makes  an  equilibrium  with  one 
of  these  partial  weights,  is  to 
this  weight,  in  the  constant 
ratio  of  the  pitch  of  the  screw, 
to  the  circumference  which 
the  power  tends  to  describe. 
Hence  the  sum  of  the  partial  weights,  or  the  total  weight 
of  the  nut,  is  to  the  sum  of  the  partial  powers,  or  to  the 
total  power  Q,  in  the  same  ratio. 

208.  From  this  it  follows :  1st.  The  force  necessary 
to  be  applied  to  the  nut  parallel  to  the  axis  of  the  screw, 
to  produce  an  equilibrium  with  the  power  Q,  which  tends 


*  This  relation  is  independent  of  the  form  of  the  fillet  of  the 
screw ;  because  the  only  question  is  to  prevent  the  moveable  piece 
from  turning  around  a  line  which  is  supposed  to  be  fixed,  No.  203. 


THE   SCREW. 


191 


to  turn  the  nut,  should  be  to  this  power,  as  the  circum 
ference  of  the  circle  which  this  power  tends  to  describe, 
is  to  the  pitch  of  the  screw ; 

2d.  For  the  same  screw,  the  effect  of  the  power  Q  is 
as  much  greater  as  this  power  is  applied  at  a  greater 
distance  from  the  axis  of  the  screw ; 

3d.  For  two  different  screws,  the  power  being  applied 
at  the  same  distance  from  the  axis,  its  effect  is  as  much 
more  considerable  as  the  height  of  the  pitch  is  less ;  that 
is  to  say,  the  closer  the  fillets  of  the  screw  are  together, 
the  greater  is  the  effect  of  the  power  for  compressing 
in  the  direction  of  the  axis. 


II. 

209.  The  screw  is 
sometimes  employed  for 
communicating  to  a 
toothed  wheel  a  motion 
of  rotation  upon  its 
axis.  For  this  purpose, 
having  given  to  the  screw 
a  pitch  DE,  equal  to  one 
of  the  divisions  of  the 
toothed  wheel,  it  is  so 
arranged  that  its  axis  is 
in  the  plane  of  the  wheel,  and  its  fillet  catches  in  the 
teeth.  This  being  done,  when  a  power  Q  turns  the  screw 
upon  its  axis,  by  means  of  a  crank  FG,  the  fillet  carries 
along  the  teeth,  which  follow  each  other,  and  it  turns 
the  wheel  in  spite  of  the  resistance  P,  which  opposes 
its  rotary  motion. 


192  STATICS. 

When  the  screw  is  employed  for  this  purpose,  it  is 
named  the  endless  screw. 

To  find  the  ratio  of  the  power  Q  to  the  resistance  P 
in  the  case  of  equilibrium,  suppose  the  resistance  is 
suspended  to  a  weight  by  a  cord  which  envelopes  the 
arbor  of  the  wheel.  By  virtue  of  this  weight,  the  tooth 
of  the  wheel  presses  the  fillet  of  the  screw  parallel  to 
the  axis  HF  ;  and  if  we  name  R  this  pressure,  we  have 
(164), 

p  :  R  : :  AC  :  AB. 

Now  we  may  regard  the  pressure  of  the  tooth  as  that 
which  a  nut,  pushed  by  a  force  R  parallel  to  the  axis 
of  the  screw,  would  exert ;  we  have,  in  the  case  of  equi 
librium, 

R  :  Q  : :  circ.  FG  :  DE  ; 

hence,  by  multiplying  together  the  two  proportions  in 
order,  we  have, 

p  :  Q  : :  ACXcirc.  Fa  :  ABXDE  ; 


that  is  to  say,  the  resistance  is  to  the  power,  as  the  pro 
duct  of  the  radius  of  the  wheel  by  the  circumference 
which  the  crank  describes,  is  to  the  product  of  the 
radius  of  the  arbor  of  the  wheel  by  the  pitch  of  the 

screw. 


THE   WEDGE. 


193 


On  the  Wedge. 

210.  The  wedge  is  a  triangular  prism 
ABCDEF,  which  is  introduced  by  its  sharp 
edge  EF  into  a  crack,  to  split  or  sepa 
rate  the  two  parts  of  a  body.  It  is 
also  made  use  of  for  exerting  great 
pressures  or  to  stretch  cords. 

Knives,  hatchets,  punches,   and,  in 
general,  all  cutting  and  penetrating  in 
struments  may  be  considered  as  wedges. 

The  face  ABCD,  upon  which  we  strike  the  wedge  to 
sink  it,  and  which  receives  the  action  of  the  power,  is 
named  the  heel  of  the  wedge;  the  side  EF,  by  which  the 
wedge  commences  to  penetrate,  is  named  the  Hade;  and 
the  name  of  sides  are  given  to  the  faces  AFED,  BFEC,  by 
which  it  compresses  the  bodies  which  it  has  to  separate  ; 
or,  since  we  are  accustomed  to  represent  the  wedge  by 
its  triangular  profile  ABF,  the  base  AB  of  the  triangle 
is  called  the  heel  of  the  wedge ;  AF  and  BF  are  its  sides. 
Fig.  96.  211.  We  are  accustomed  to  suppose 

that  the  direction  of  the  power  is  per 
pendicular  to  the  heel  of  the  wedge ; 
because,  generally,  the  wedge  is  slink  by 
striking  upon  the  heel  with  a  hammer, 
or  with  any  other  object  which  has  no 
connection  with  it,  and  which,  in  this 
case,  if  the  direction  CD  of  the  shock  is 
not  perpendicular  to  the  surface  of  the 
heel,  the  action  is  naturally  decomposed  into  two  other 
forces,  CH,  CE  ;  the  first  of  which,  being  parallel  to  the 


194 


STATICS. 


heel  of  the  wedge,  can  have  no  other  effect  than  that 
of  making  the  hammer  slip,  and  the  second  being  per 
pendicular  to  the  face  AB,  is  the  only  one  which  is  trans 
mitted  to  the  wedge,  and  conduces  to  the  effect  which 
is  to  be  produced.  But  if  the  power  were  applied  to 
the  wedge  by  means  of  a  cord,  whose  point  of  attach 
ment  could  not  slip,  then,  by  considering  this  power,  it 
would  be  necessary  to  take  account  of  its  direction. 


I. 

212.  Let  c,  D,  be  two 
points  separated  by  a 
wedge  ABF,  and  retained 
by  a  cord  CD,  which  is 
attached  to  them,  and 
which  opposes  their  sepa- 

j  V''''"|\  •*••    /T\j         "  rati°n?    suppose,  more- 
*•£>     ™   ••»  A'""^-       over,  that  these  points 

are  supported  against  a 
resisting  plane  which 
prevents  them  from  moving  in  a  direction  perpendicu 
lar  to  the  cord,  and  that  a  power  P,  applied  perpendicu 
larly  to  the  heel  AB  of  the  wedge,  makes  an  effort  to 
separate  them  and  move  them  in  the  direction  of  the 
length  of  this  cord.  This  being  established,  it  is  re 
quired  to  determine,  1st,  the  tension  which  results  in 
the  cord  CD  ;  2d,  the  force  necessary  to  be  applied  to 
one  of  the  two  points  c,  D,  to  prevent  them  from  moving 
in  the  direction  of  the  cord;  3d,  the  pressure  which 
each  of  these  points  exerts  upon  the  plane  which  resists 
them. 


THE   WEDGE.  195 

It  must  be  remarked  first,  that,  if  the  direction  of  the 
power  P  is  not  such  that  it  may  be  decomposed  into  two 
others  Q,  R,  whose  directions  pass  through  the  points  i 
c,  D,  and  are  perpendicular  to  the  sides  AF,  BF,  the  ( 
wedge  will  turn  between  the  two  points  c,  D  until  this  , 
condition  is  fulfilled,  and  then  only  will  the  power  P  j 
produce  all  its  effect.    We  will  suppose,  moreover,  that,  ' 
having  drawn  through  the  points  c,  D  the  lines  CE,  DE»* 
perpendicular  to  the  sides  of  the  wedge,  the  point  of 
intersection  E  of  these  two  lines  is  in  the  direction  of 
the  power  P. 

This  being  the  case,  the  force  P  will  be  decomposed 
into  two  other  forces  Q,  R,  directed  along  EC,  ED  ;  and, 
if  we  represent  this  force  by  the  part  EX  of  its  direc 
tion,  and  finish  the  parallelogram  EZXY,  we  shall  have, 

p  :  Q  :  R  : :  EX  :  EY  :  EZ  or  YX ; 

or,  since  the  two  triangles  EXY  and  ABF,  whose  sides  are 
perpendicular  each  to  each,  are  similar,  we  shall  have, 

p  :  Q  :  R  : :  AB  :  AF  :  BF, 
and,  consequently, 


PXAF  PXBF 

Q= ,  R= , 

AB  AB 


The  point  c  not  being  able  to  move  in  the  direction 
ECH,  because  of  the  resistance  of  the  plane  upon  which 
it  is  supported,  the  force  Q,  which  is  applied  to  it,  will 
be  decomposed  into  two  others ;  one  of  which,  in  the 


196 


STATICS. 


direction  of  the  line  ci,  perpendicular  to  the  cord,  will 
be  destroyed  by  the  resistance  of  the  plane,  and  the 
other,  in  the  direction  of  the  prolongation  of  DC,  will 
be  employed  in  stretching  the  cord.  Thus,  by  making 
CH=EY,  and  completing  the  rectangle  CGHI,  the  two 
components  of  the  force  Q  will  be  represented  by  ci  and 
CG  ;  and  we  shall  have, 

Q  :  force  ci  :  force  CG  : :  CH  :  ci  :  CG, 
and,  consequently, 


force  ci 


QXCI 


»  QXCG 

force  CG= : 

CH 


or,  substituting  for  Q  its  value  previously  found,  we  shall 
have, 

PXAFXCI 


force  ci= 


ABXCH 

PXAFXCG 

ABXCH 

In  like  manner,  if, 
upon  the  prolongation 
of  ED,  we  make  DL=EZ, 
and  complete  the  rect 
angle  DKLM,  whose  side 
DK  is  upon  the  prolonga 
tion  of  CD,  and  whose 
side  DM  is  perpendicular 
to  CD,  the  force  E  will 
be  decomposed  into  two 


THE   WEDGE.  197 

others  DM,  DK,  the  first  of  which  will  be  destroyed  by 
the  resistance  of  the  plane,  and  the  second  will  be 
wholly  employed  in  acting  upon  the  cord ;  and  we  will 
find,  likewise, 


„  RXDM      PXBFXDM 

force  DM= =—     — , 


force  DK= 


DL  ABXDL 

RXDK      PXBFXDK 


DL  ABXDL 


Thus  the  cord  CD  will  be  drawn  in  one  direction  by  the 
force  CG  and  in  the  contrary  direction  by  the  force  DK. 

Now,  when  a  cord  is  drawn  in  contrary  directions  by 
two  unequal  forces,  the  tension  which  it  suffers  is  always 
equal  to  the  smaller  of  these  two  forces ;  for,  when  the 
two  forces  are  equal,  one  of  them  is  the  measure  of  the 
tension  of  the  cord;  and  when  they  are  unequal,  the 
excess  of  the  greater  over  the  smaller,  not  being  coun 
ter-balanced  by  anything,  does  not  contribute  to  stretch 
the  cord,  and  has  no  other  effect  than  to  draw  it  along 
in  the  direction  of  its  length. 

Hence,  1st,  the  tension  of  the  cord  CD  will  be  equal 
to  the  smallest  of  the  two  forces  CG,  DK. 

2d.  The  cord  will  be  drawn  along  in  the  direction  of 
its  length  and  in  that  of  the  greater  of  the  two  forces 
CG,  DK  ;  so  that,  in  order  to  oppose  this  motion,  it  will 
be  necessary  to  apply  to  one  of  the  two  points  c,  D  a 
force  equal  to  the  difference  of  these  two  forces,  and 
directly  opposed  to  the  greater. 

3d.  The  pressures  exerted  by  the  two  points  c,  D 
upon  the  plain  which  retains  them,  will  be  equal ;  the 
first  to  the  force  ci,  the  second  to  the  force  DM. 

17* 


198  STATICS. 


II. 

fr-  98-  213.  If  the  sides  AF,  BF 

p|  of  the  wedge  be  equal,  the 

heel  AB  is  parallel  to  the 
cord  which  retains  the  two 
points  c,  D  ;  and,  at  the  same 
time,  if  the  direction  of  the 
force  P  is  perpendicular  to 
the  middle  of  AB,  1st,  the 
wedge  will  not  turn,  because  the  lines  CE,  DE,  drawn 
through  the  two  points  of  support  perpendicular  to  the 
sides  of  the  wedge,  will  meet  in  a  point  E  of  the  di 
rection  of  the  power.  2d.  Both  sides  being  perfectly 
alike,  the  forces  CG,  DK  will  be  equal  to  each  other,  and 
each  of  them  will  be  the  measure  of  the  tension  of  the 
cord  CD.  3d.  By  letting  fall  from  the  point  F  the  per 
pendicular  FN  upon  the  heel  of  the  wedge,  the  two  tri 
angles  CGH,  BNF,  whose  sides  will  be  perpendicular  each 
to  each,  will  be  similar,  and  give, 

CH  :  CG  : :  BF  :  FN. 
Now  we  have, 

Q  :  force  CG  : :  CH  :  CG, 

and,  consequently, 

Q  :  force  CG  : :  BF  :  FN  ; 

But,  we  have,  also, 

p  :  Q  : :  AB  :  BF. 


I 

THE    WEDGE.  199 

Hence,  by  multiplying  together  these  two  proportions 
in  order,  we  shall  have 

p  :  force  CG  : :  AB  :  FN  ; 

that  is  to  say,  in  this  case,  the  power  p  will  be  to  the 
tension  of  the  cord  CD,  as  the  heel  of  the  wedge  is  to 
its  height. 

We  will  not  make  any  application  of  the  theory  of 
the  wedge  to  the  use  which  may  be  made  of  this  in 
strument  for  splitting  bodies  ;  because,  under  such  cir 
cumstances,  the  resistance  which  has  to  be  overcome  is 
always  unknown,  and  it  is  useless  to  know  the  ratio  of 
this  resistance  to  the  power  which  is  in  equilibrium 
with  it. 


LEMMA. 

214.  If  from  the  summits  B,  c  of  two  of  the  angles 
of  a  triangle,  the  perpendiculars  BE,  CD  be  dropped  upon 
the  opposite  sides,  these  perpendiculars  will  be  to  each 
other  reciprocally  as  the  sides  upon  which  they  are 
dropped  ;  that  is  to  say,  we  shall  have, 

BE  :  CD  : :  AB  :  AC. 

DEMONSTRATION.  By  con 
sidering  AB  as  the  base  of 
the  triangle,  the  perpen 
dicular  CD  will  be  its  height, 
and  the  surface  of  the  tri- 

i          »TI     -1         AI>  X  GD        -r- 

angle  will  be   -— = — .     In 


200  STATICS. 

like  manner,  by  taking  AC  for  the  base,  the  surface  will 

be  -       —  ;  hence,  we  shall  have  the  two  equal  products 
& 

ACXBE=ABXCD,  which  will  give  the  proportion, 
BE  :  CD  : :  AB  :  AC. 


THEOREM. 

215.  When  the  two  powers  P,  Q  are  applied  to  the 
faces  AB,  AC  of  a  wedge,  the  third  face  of  which  EC  is 
supported  upon  a  resisting  plane  MN,  and  these  two 
powers  are  in  equilibrium  through  the  resistance  of  the 
plane,  they  are  to  each  other  reciprocally  as  the  spaces 
which  they  traverse  in  the  line  of  their  directions,  if  the 
equilibrium  be  disturbed  infinitesimally. 
Fig.  100. 


DEMONSTRATION.  Since  the  powers  P,  Q  are  in  equi 
librium,  their  directions  are  perpendicular  to  the  faces 
of  the  wedge  to  which  they  are  applied  (179).  Now 
suppose,  by  virtue  of  a  derangement  in  the  equilibrium, 
the  wedge  slips  upon  the  resisting  plane,  and  takes  the 
infinitely  near  position  abc ;  and  the  direction  QE  is  pro 
longed  until  it  meets  ac  in  e :  it  is  evident  that  the 


THE    WEDGE.  201 

small  lines  Ee,  Dd  will  be  the  spaces  which  the  powers 
will  have  traversed  in  the  lines  of  their  directions. 

Lastly,  draw  Aa ;  prolong  CA,  ba  until  they  intersect 
somewhere  in  F,  and  from  the  points  A,  a  let  fall  the 
perpendiculars  AH,  aa ;  we  will  have  evidently  Ee=Ga 
and  j)d=AH. 

This  being  determined,  the  powers  P,  Q  being  in  equi 
librium,  they  are  to  each  other  as  the  sides  of  the 
wedge  to  which  they  are  applied ;  then  we  shall  have 
(212), 

p  :  Q  : :  AB  :  AC, 
and,  because  of  the  similar  triangles  ABC,  F«A, 

p  :  Q  : :  F#  :  FA  ; 
hence  (214),  we  shall  have, 

p  :  Q  : :  a&  :  AH 
and,  consequently, 

p  :  Q  : :  Ee  :  Dd. 

216.  We  have  seen  that  the  analogous  proposition 
takes  place  in  the  equilibrium  of  all  the  machines  which 
we  have  considered.  By  following  the  same  process,  it 
could  be  demonstrated  directly  that,  when  two  powers 
are  in  equilibrium  by  means  of  points  of  support  which 
any  machine  presents,  they  are  to  each  other  reciprocally 
as  the  spaces  which  they  would  traverse  in  the  lines  of 
their  directions,  if  the  equilibrium  were  infinitesinr.ally 


202  STATICS. 

disturbed.  By  means  of  this  proposition,  it  will  be 
easy  to  find  in  practice  the  relation  which  should  subsist 
between  a  power  and  a  resistance  applied  to  a  proposed 
machine,  in  order  that  these  forces  should  be  in  equi 
librium,  leaving  out  of  consideration  friction  and  other 
obstacles  to  motion. 


203 


NOTE. 


A  NEW  DEMONSTRATION  OF  THE  PARAL 
LELOGRAM  OF  FORCES. 

BY   M.    CAUCHY.* 

IF  we  suppose,  as  in  general,  a  force  to  be  repre 
sented  by  a  length  laid  off  from  its  point  of  application 
along  the  direction  in  which  it  acts,  the  resistance  R  of 
two  forces  P,  Q,  simultaneously  applied  to  a  material 
point  (A),  will  be  represented  in  intensity  and  direction 
by  the  diagonal  of  the  parallelogram  constructed  upon 
these  two  forces.  This  proposition  has  already  been 
demonstrated  in  several  manners.  But,  among  the 
demonstrations  which  have  been  given,  some  require  the 
consideration  of  new  material  points  connected  with  the 
point  (A)  by  rigid  and  invariable  lines ;  others  are 
founded  upon  the  use  of  the  differential  calculus,  or  of 
derived  functions;  others  again  are  deduced  from  the 
relations  which  exist  between  the  cosines  of  certain 
angles.  I  am  here  about  to  demonstrate  the  same  propo 
sition  without  recurring  to  these  different  considerations, 

*  This  demonstration  is  extracted  from  the  work  -which  M.  Cauchy 
publishes  in  numbers  under  the  name  of  Exercises  de  Mathematiques. 


204  STATICS. 

and,  in  order  to  attain  my  object,  I  will  establish  suc 
cessively  several  lemmas,  which  may  be  enunciated  as 
follows. 


LEMMA  I. 

If  we  designate  by  R  the  resultant  of  the  two  forces 
p,  Q,  simultaneously  applied  to  the  point  (a),  and  by  x 
any  number,  the  resultant  of  two  forces,  equal  to  the  pro 
ducts  P#,  Qx,  and  directed  along  the  same  lines  as  the 
forces  P,  Q,  will  be  represented  by  the  product  RZ,  and 
directed  along  the  same  line  as  the  force  R. 

DEMONSTRATION.  In  the  first  place  let  #=— ,  m  and 
n  representing  any  two  whole  numbers.  We  shall  have 

_     p  Q 

n  n 

Besides,  we  may  consider  the  component  P,  or  m  —,  as 

n 

produced  by  the  addition  of  several  forces  equal  to  -, 

n 

and  the  component  Q,  or  m  - ,  as  produced  by  the  addi- 

n 

tion  of  as  many  forces  equal  to  — .    From  this  it  is  easy 

to  conclude,  that  the  resultant  of  the  forces  P#,  Q#,  will 
both  be  produced  by  the  addition  of   several   forces 

p        o 

equal  to  the  resultant  of  -  and  -.     Moreover,  it  is  evi- 
n         n 

dent  that  the  first  two  resultants  are  to  the  last,  as  the 


NOTE.  205 

numbers  n  and  m  are  to  unity.  Hence,  the  second  re 
sultant  will  be  equivalent  to  the  first  multiplied  by  the 

997 

ratio  —=Xj  that  is  to  say,  to  the  product  nx. 

% 

Let  us  suppose  in  the  second  place  that  the  number 
x  is  irrational.     Then  we  can  vary  the  whole  numbers 

//77 

m  and  n  in  such  a  way  that  the  fraction  —  will  converge 

7l/ 

to  the  limit  x;  and  it  is  evident  that,  in  this  case,  the 
resultant  of  the  forces  — ,  — ,  always  directed  along 

the  same  line,  and  always  equal  to  — ,  will  tend  more 

n 

and  more  to  coincide,  in  intensity  and  direction,  with  the 
resultant  of  the  forces  PX,  Q#.  Hence,  this  last  result 
ant  will  be  directed  along  the  same  line  as  the  force  R, 
and  it  will  have  for  measure  the  limit  of  the  product 

— ,  that  is  to  say,  the  product  ~RX. 
n 


w 

COROLLARY. 

A 

If  we  designate  by  the  notation  (p,  Q)  the  angle  in 
cluded  between  the  directions  of  the  two  forces  P  and  Q 

A        A 

(p,  R),  (Q,  R)  will  be  the  angles  included  between  the  di 
rections  of  the  components  P  and  Q  and  their  resultant 
R.  This  being  fixed,  if  we  make  successively  R#=P, 

P         Q 
R#=Q,  or,  what  amounts  to  the  same,  #=-,  #=-,  we 

R  R 

18 


206  STATICS. 


ill  conclude  from  the  preceding  lemma,  1st,  that  the 

p2  pQ 

force  P  may  be  replaced  by  two  components  —  and  —  , 

Ei  11 

A  A 

which  form  with  it  angles  equal  to  (P,  R)  and  (Q,  K)  ; 

2d,  that  the  force  Q  may  be  replaced  by  two  com- 

O2  PQ 

ponents  —  and  —  ,   which  form  with  it  the  angles  (Q,  R) 

R  II 

A 
and  (P,  R). 


LEMMA  II. 

The  resultant  R  of  two  forces  P,  Q,  which  intersect  at 
right  angles*  is  represented  in  intensity  by  the  diagonal 
of  the  triangle  constructed  upon  the  two  components,  so 
that  we  have, 


DEMONSTRATION.     Let  us  conceive  the  force  P  to  be 
replaced  by  the  two  above-mentioned  components,  that 


"P  "PO 

is   to   say,  by   two   forces  —  and  — ,  which  form  with 

R  R 

it  the  angles  (P,  R)  and  (Q,  R).    Let  us  conceive  also  that 

Q2        i   PQ 
the  force  Q  is  replaced  by  two  components  -  and  — , 

A  A 

which  form  with  it  the  angles  (Q,  R)  and  (P,  R).     We 


NOTE.  207 

p2     Q2 

can  suppose  that  the  forces  — ,  —  are  directed  along  the 

R      R 

same  line  as  the  resultant  R,  and  then  the  two  forces 

PO 
equivalent  to  —  will  each  form  with  the  direction  of  R 

an  angle  equal  to  (P,  Q).  Hence,  they  will  form  between 

A 
them  an  angle  equal  to  double  (P,  Q).     Hence,  since  the 

A 
angle  (P,  Q)  is  a  right  angle,  by  hypothesis,  the  forces 

PQ 
equivalent  to  —  will  be  equal,  but  directly  opposed. 

• 

Consequently,  they  will  be  in  equilibrium ;  and  for  the 

p2     Q2 

forces  — ,  — ,  directed  along  the  same  line,  we  can  sub- 
R    n 

stitute  only  the  forces  P,  Q,  or  their  resultant  R.    Hence, 
we  shall  have  the  equation, 

P2  ,  Q2 

R=— +— , 

R     R' 

from  which  formula  (1)  is  immediately  deduced. 
This  demonstration  is  due  to  Daniel  Bernoulli. 


LEMMA  III. 

The  resultant  R  of  the  two  forces  P,  Q,  which  intersect 
at  right  angles,  is  represented,  not  only  in  intensity,  as 
we  have  proved  above,  but  also  in  direction,  by  the  diago 
nal  of  the  rectangle  constructed  upon  the  two  components. 

DEMONSTRATION.  This  proposition  is  evident  in  the 
case  where  the  forces  P,  Q  are  equal  to  each  other. 


208  STATICS. 

Then  the  resultant  R  should  necessarily  divide  the  angle 

A 
(p,  Q)  into  two  equal  parts,  and  we  have,  by  virtue  of 

Lemma  II, 

R2=2p2,  or  R=pv/2. 

It  is  also  easy  to  demonstrate,  Lemma  III,  in  the  case 
where  we  suppose  Q2=2p2,  or  Q=p^2.  Thus,  we  will 
consider  three  forces,  equal  to  P,  directed  along  three 
lines  which  are  perpendicular  to  each  other.  These 
three  forces  will  be  represented  by  three  sides  of  a  cube 
which  meet  in  the  same  summit.  Moreover,  the  result 
ant  of  two  of  these  forces  being  equal  to  p^2,  and  di 
rected  along  the  diagonal  of  one  of  the  forces  of  the 
cube,  the  resultant  Q  of  the  three  forces  will  of  necessity 
be  included  in  the  whole  plane,  which  will  contain  one 
of  the  forces  P  and  the  diagonal  of  the  square  con 
structed  upon  the  other  two.  Now,  there  are  three 
planes  of  this  kind,  and  these  three  planes  intersect  in 
the  diagonal  of  the  cube.  Hence,  the  resultant  of  the 
three  forces  P,  or  what  is  the  same,  the  resultant  of  the 
forces  P  and  P^S,  which  intersect  at  right  angles,  will 
be  directed  along  the  diagonal  of  the  cube,  which  is  at 
the  same  time  the  diagonal  of  the  rectangle  constructed 
upon  the  forces  P  and  p^2. 

It  might  be  proved,  precisely  in  the  same  manner, 
that  if  we  designate  by  m  a  whole  number,  and  suppose 
Lemma  III  to  be  demonstrated  in  the  case  wrhere  we 
have  Q— p^w,  the  resultant  of  three  forces,  respect 
ively  equivalent  to 

p,  P, 


NOTE.  209 

and  represented  by  three  lines  perpendicular  to  each 
other,  will  be  directed  along  the  diagonal  of  the  rect 
angular  parallelopipedon,  which  will  have  these  same 
lines  for  sides.  From  this  we  conclude,  in  the  admitted 
hypothesis,  that  Lemma  III  will  also  subsist,  if  we  take 
for  Q  the  resultant  of  the  forces  P  and  P^w,  that  is  to 
say,  if  we  make  Q=p^m+l.  Besides,  Lemma  III  is 
evident,  when  we  have  Q=P,  or,  what  is  the  same,  m=1. 
Hence,  this  lemma  will  also  subsist,  if  we  take 
Q=P\/l-f  l==pv/2,  or  Q=pN/2+l=p>/3,  etc., 

or,  in  general,  Q=P^/W,  m  being  any  whole  number. 

Let  us  conceive  now,  that  m  and  n  designate  two 
whole  numbers  ;  and  construct  a  rectangular  parallelo 
pipedon,  which  has  for  its  sides  three  lines  representing 
the  three  forces 


The  resultant  of  these  three  forces  will  evidently  be 
contained;  1st,  in  the  plane  which  includes  the  force 
p^n  and  the  diagonal  p^m+1  of  the  rectangle  con 
structed  upon  the  forces  P,  p^m,  2d,  in  the  plane 
which  includes  the  force  p^m  and  the  diagonal 
of  the  rectangle  constructed  upon  the  forces  P, 
Hence,  this  resultant  will  be  directed  along  the  diagonal 
of  the  parallelopipedon  ;  and  the  plane,  which  contains 
the  same  resultant  with  the  force  P,  will  intersect  the 
plane  of  the  two  forces  p^m,  P^n  along  the  diagonal 
of  the  rectangle  constructed  upon  these  two  forces. 
Hence  the  resultant  of  the  forces  ?*/m,  p-v/w,  which 
evidently  should  be  comprised  in  the  plane  in  question, 

18* 


210  STATICS. 

will  be  directed  along  this  last  diagonal.  Hence  Lemma 
III  will  subsist,  when  we  replaee  the  forces  P  and  Q 
by  two  forces  equal  to  Q^w,  P^n,  that  is  to  say;  by 
two  forces  whose  squares  are  to  each  other  in  the  ratio 
of  m  to  n.  Hence,  Lemma  III  will  also  subsist  between 
the  forces  P  and  Q,  if  we  suppose 


;=pj~- 


Now,  let  Q=Payx  designating  any  number  whatever. 
We  can  vary  the  whole  numbers  m  and  n,  in  such  a 

manner  that  the  ratio  —  will  converge  to  the  limit  ar, 
n 

and  it  is  evident,  that,  in  this  case,  the  resultant  of  the 
forces  P  and  P*J  — ,  directed  along  two  lines  perpendicu 
lar  to  each  other,  will  tend  more  and  more  to  coincide, 
in  intensity  and  direction,  on  the  one  hand  with  the  re 
sultant  of  the  forces  P,  PX,  and,  on  the  other,  with  the 
diagonal  of  the  rectangle  constructed  upon  these  two 
forces.  Hence,  the  resultant  of  the  forces  P,  ?x  will 
be  represented  by  the  diagonal  in  question. 


COROLLARY  I. 

If  the  force  R  be  represented  by  the  length  AB  laid 
off  from  its  point  of  application  (A)  upon  the  line  along 
which  it  acts,  and  if  we  draw  through  the  point  (A)  two 
axes  perpendicular  to  each  other,  we  may  substitute  for 
the  force  R  the  two  forces  represented  in  intensity  and 


NOTE.  211 

direction  by  the  projections  of  the  line  AB  upon  the 
two  axes. 


COROLLARY  II. 

Let  us  conceive  now,  that  two  forces  P,  Q,  being  applied 
to  the  same  point  (A),  and  represented  by  two  lines  AB, 
AC,  which  form  any  angle  between  them,  to  be  traced  in 
the  plane  of  these  two  forces,  two  axes,  one  of  which 
coincides  with  the  diagonal  of  the  parallelogram  to 
which  they  serve  as  sides,  and  the  other  perpendicular 
to  this  diagonal.  We  can  substitute  for  the  two  forces 
p,  Q,  the  four  forces  represented  in  intensity  and  di 
rection  by  the  projections  of  the  lines  AB,  AC  upon  the 
two  axes.  Now,  of  these  four  forces,  two,  being  directly 
opposed,  will  be  in  equilibrium.  The  other  two,  directed 
along  the  diagonal  of  the  parallelogram,  will  be  added 
together,  and  will  give,  for  their  sum,  a  force  represented 
in  intensity  and  direction  by  this  same  diagonal.  Hence 
we  may  enunciate  the  following  proposition : 


THEOREM. 

The  resultant  R  of  two  forces  P,  Q  simultaneously  ap- 

:  plied  to  a  material  point  (A)  and  directed  in  any  manner 

whatever,  is  represented  in  intensity  and  direction  by 

the  diagonal  of  the  parallelogram  constructed  upon  these 

two  forces. 


212  STATICS. 


COROLLARY  I. 

As  the  diagonal  R  of  the  parallelogram  constructed 
upon  the  two  forces  P,  Q  is  at  the  same  time  the  third 
side  of  the  triangle,  which  is  formed  by  drawing  through 
the  extremity  of  the  first  force  a  line  equal  and  parallel 
to  the  second,  and  as  the  angle  in  this  triangle,  opposite 

A 
to  the  side  R,  is  the  supplement  of  the  angle  (P,  Q),  we 

have  necessarily,   by  virtue  of   a  known   formula   in 
trigonometry, 

A 

R2  =  p2-f  Q2  +  2PQ  COS   (P,  Q) (2). 


COROLLARY  II. 

In  the  case  where  the  forces  P,  Q  become  equal  to 
each  other,  their  resultant  R  is  represented  in  intensity 
and  direction  by  the  diagonal  of  the  lozenge  constructed 
upon  these  same  forces.  Then  the  formula  (2)  re 
duces  to 

A 

R2=2P2  fl  +  COS(P,  Q)J.         (3) 

A 
Moreover,  by  supposing  (P,  R)=0,  we  will  find,  in  the 

A 
present  case,  (P,  Q)=0;  and  as  we  have,  in  general, 

cos  2^=2  cos2  o—l,          (4) 

the  equation  (3)  will  give  R=2p  cos  0,  or,  what  is   the 
same, 

A 
R=2p  cos  (P,  R).  (5) 


NOTE.  213 

For  the  rest,  we  may  be  directly  assured  that  the 
second  member  of  the  formula  (5)  represents  the  di 
agonal  of  the  lozenge  constructed  upon  the  two  forces 
equal  to  P. 

It  is  easy  to  demonstrate  the  theorem  of  page  211, 
for  the  case  in  which  the  forces  P,  Q  have  any  ratio  to 
each  other,  when  we  have  once  established  this  theorem 
for  the  case  where  we  have  Q=P,  that  is  to  say,  when 
formula  (5)  is  established.  Now,  we  can  give  a  direct 
demonstration  of  this  formula,  which,  in  fact,  is  deduced 
from  equation  (4),  but  which  appears  to  merit  special 
remark.  I  will  explain  it  in  a  few  words. 

Let  us  admit  that  formula  (5)  is  verified  for  the  case 
A 

where  we  have  (P,  R)— *,  *  designating  either  a  right 
angle  or  an  acute  angle.  I  say  that  it  will  still  subsist, 
if  we  suppose 

(p,n)=|,  or(p,R)=?-|. 

A 
Now,  in  these  two  hypotheses,  the  angle  (P,  Q),  included 

between  the  directions  of  the  two  equal  forces  P,  Q, 
will  be  equivalent  to  one  of  the  angles  *,  rt—t;  and  we 
can  prove  by  reasoning  as  in  Lemma  II,  that  we  can 
substitute  in  the  system  of  the  two  forces  P,  Q,  or  for 

p2 

their  resultant  R,  four  components  equal  to  — ,  among 

which,  two  will  be  directed  along  the  same  line  and  in 
the  same  direction  as  the  force  R,  while  the  other  two 
will  each  form  with  the  resultant  R  an  angle  equivalent 

A 
to  (P,  R),  that  is  to  say,  to  *  or  to  *— *.    Now,  since  we 


214  STATICS. 

suppose  formula  (5)  to  be  verified  in  the  case  where  we 

A 
have  (P,  R)=*,  the  last  two  components  evidently  can 

p2 

be  replaced  by  a  single  force  equal  to  2—  cos  *,  and 

R 

situated  in  the  direction  of  the  resultant  R,  or  in  the 
opposite  direction.  Consequently,  we  will  find  defini 
tively, 

R=2-±2-  cos  *=2-  (Idbcos  *) ; 
R        R  R  v 

or, 

R2=2P2  (1±COS  *),  (6) 

the  sign  ±  being  necessarily  reduced  to  the  sign+,  in 
the  case  where  (P,  R)=O?  and  to  the  sign  —  in  the  case 

where  (P,  R)=^— -.  Since  we  will  also  get  the  formula 
(4)  by  placing  successively 


1+cos  *=2  cos2  5,  1— cos  *=2  cos2  — — , 
equation  (6)  will  give  in  the  first  case 

R=2  cos  o, 
and,  in  the  second, 


7i  —  if 
R  =  COS 


2 
Hence,  if  equation  (4)  subsists  when  we  attribute  to  the 

angle  (P,  R)  the  value  t,  it  will  also  subsist  when  we 


NOTE.  215 

-.  ! 

2       i 


attribute  to  the  same  angle  one  of  the  values  ~, 


Now,  this  equation  is  verified  when  we  suppose  (P,  R)=5, 

jj 

since  we  have  evidently  in  this  hypothesis  R=o,  and 

A 
cos  (P,  R)  =0.     Hence,  it  will  be  equally  true  if  we 

suppose 


and,  consequently,  if  we  take 

/  A  N     1   *    n          A       i         -      3^ 

<P>  B)=2'  4=8'  or(p>R)=2  (*-3=-s  ' 

Hence  it  will  also  be  true,  if  we  attribute  to  the  angle 

A 
(p,  E)  one  of  the  values, 

1     7t  _  *     1    07t  _  3rt  -lx        3rtv  _  5rt 

2'8~16'  2'  8""l6'  2^~8"^"~16' 


By  continuing  in  the  same  manner,  it  may  be  proved 

A 
that  formula  (5)  generally  has  place  when  the  angle  (P,  R) 

2?w4-l 
receives  a  value  of  the  form  -  -,  n  representing  any 

whole  number,  and  2m  +1  an  odd  number  less  than  2n. 
If  we  now  represent  by  0  an  acute  angle  taken  at  will, 
we  can  vary  the  whole  numbers  m  and  n  in  such  a  man 

ner  that  the  ratio  —  =  —  will  indefinitely  approach  the 


216  STATICS. 

limit  e ;  and  the  resultant  n  will  tend  more  and  more 
to  coincide,  on  the  one  hand,  with  a  force  equivalent  to 
2p  cos  0,  and,  on  the  other  hand,  with  the  resultant  of 
two  forces  equal  to  P,  which  would  form  between  them 
an  angle  double  of  o.  Hence,  this  last  resultant  will 
be  represented  in  magnitude  by  2p  cos  0,  and  will  also 
verify  formula  (5). 


THE    END. 


STEREOTYPED  BY  L.  JOHNSON  AND  CO. 
THILADELrHIA. 


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